{"id":89,"date":"2018-10-29T11:39:07","date_gmt":"2018-10-29T11:39:07","guid":{"rendered":"http:\/\/mgmtp15.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=89"},"modified":"2022-01-07T05:28:20","modified_gmt":"2022-01-07T05:28:20","slug":"probability-theory-concept-and-enumeration","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/chapter\/probability-theory-concept-and-enumeration\/","title":{"rendered":"Probability Theory: Concept And Enumeration"},"content":{"raw":"
\r\n\r\n \r\n\r\n1.\u00a0\u00a0\u00a0\u00a0\u00a0 Introduction\r\n\r\n \r\n\r\n2.\u00a0\u00a0\u00a0\u00a0\u00a0 Terminology\r\n\r\n \r\n\r\n3.\u00a0\u00a0\u00a0\u00a0\u00a0 Properties of Probability\r\n\r\n \r\n\r\n4.\u00a0\u00a0\u00a0\u00a0\u00a0 Approaches To Probability\r\n\r\n \r\n\r\n5.\u00a0\u00a0\u00a0\u00a0\u00a0 Problems\r\n\r\n \r\n\r\n6.\u00a0\u00a0\u00a0\u00a0\u00a0 Summary\r\n\r\n \r\n\r\n \r\n\r\nIntroduction<\/strong>\r\n\r\n \r\n

Probability which measures the likelihood that an event will occur is an important part of statistics. It is the basis of inferential statistics Here we make decisions under conditions of uncertainty. Probability theory is used to evaluate the uncertainty involved in those decisions. The concept of statistical probability may be characterized by saying that probability do not refer to judgment but predict the possible outcomes of a conceptual experiment, such as tossing a coin or counting the number of telephone calls, estimating next year\u2019s sales for a company is based on many assumptions, some of which may happen to be true and others may not.<\/p>\r\n \r\n

Probability theory will help us make decisions under such conditions of imperfect information and uncertainty. Combining probability & probability distribution with descriptive statistics will help us make decision about population based on information obtained from samples.<\/p>\r\n \r\n\r\nTERMINOLOGY:<\/strong>\r\n\r\n \r\n\r\nRANDOM EXPERIMENT:<\/strong>\r\n\r\n \r\n

Any experiment performed under normal circumstances which can be repeated under the same conditions and which results in one or only one of many observations. These observations are called outcome of the experiment.<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nOUTCOME:<\/strong>\r\n\r\n \r\n\r\nEach element of a sample space is called an outcome. It is also called a sample point.\r\n\r\n \r\n\r\nSAMPLE SPACE:<\/strong>\r\n\r\n \r\n

The set of all possible outcomes of a trial (or experiment) is called a sample space and is generally denoted by S.<\/p>\r\n \r\n\r\nEXPERIMENT, OUTCOME & SAMPLE SPACE:<\/strong>\r\n\r\n \r\n\r\nSample space for a weather forecaster to predict the rain is written as\r\n\r\n \r\n\r\nS = {rainy day, dry day}<\/em>\r\n\r\n \r\n\r\nThe elements of a sample space are called sample point.\r\n\r\n \r\n\r\nSample space for a student to get success or failure in semester is written as S = {Fail, pass}<\/em>\r\n\r\n \r\n\r\nLists of few examples of experiment, outcome & Sample space.\r\n\r\n \r\n\r\nTable:\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Experiment<\/strong><\/td>\r\nOutcomes<\/strong><\/td>\r\nSample space<\/strong><\/td>\r\n<\/tr>\r\n
Toss a coin<\/td>\r\nHead, Tail<\/td>\r\nS = {Head, Tail}<\/td>\r\n<\/tr>\r\n
Roll a dice<\/td>\r\n1,2,3,4,5,6<\/td>\r\nS = {1, \u2026\u2026..6}<\/td>\r\n<\/tr>\r\n
Toss a coin twice<\/td>\r\nHH, HT, TH, TT<\/td>\r\nS = {HH, HT, TH, TT}<\/td>\r\n<\/tr>\r\n
Birth of a baby<\/td>\r\nBoy, Girl<\/td>\r\nS = {Boy, Girl}<\/td>\r\n<\/tr>\r\n
Take a test<\/td>\r\nPass, Fail<\/td>\r\nS = {Pass, Fail}<\/td>\r\n<\/tr>\r\n
Select a student<\/td>\r\nMale, Female<\/td>\r\nS = Male, Female<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe sample space for an experiment can be understood by drawing a venn diagram as well as a tree diagram let us take examples.\r\n\r\n \r\n\r\nTOSSING A COIN ONCE:<\/strong>\r\n\r\n<\/div>\r\n
\r\n\r\n\"\"\"\"\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nTherefore, the sample space for two tosses of the coin is\r\n\r\n \r\n\r\nS = {HH, HT,\u00a0 TH,\u00a0 TT}\r\n\r\n \r\n

Example 1 : Suppose we randomly select two babies from a hospital and observe whether a baby selected each time is a baby boy or a baby girl. Write all outcomes for this experiment.<\/p>\r\n\"\"\r\n\r\nEQUALLY LIKELY EVENTS :<\/strong>\r\n\r\n \r\n\r\nThe event are said to be equally likely when one event does not occur more often than the other.\r\n\r\n \r\n\r\nEVENT<\/strong>\r\n\r\n \r\n\r\nAn event is a collection of one or more of the outcomes of an experiment.\r\n\r\n \r\n\r\nEVENT:<\/strong>\r\n\r\n \r\n

An event is a collection of one or more of the outcomes of an experiment. E.g. tossing of a coin is trial and getting Head (H) or Tail (T) is an event.<\/p>\r\n \r\n\r\nFAVOURABLE CASES:<\/strong>\r\n\r\n \r\n

The number of cases favorable to an event in any trial is the number of outcomes which entail the happening of the event.<\/p>\r\n \r\n\r\nEXHAUSTIVE EVENTS:<\/strong>\r\n\r\n \r\n

The total number of possible outcomes in any trial is called exhaustive events if taken together covers the sample space.<\/p>\r\n \r\n\r\nMUTUALLY EXCLUSIVE EVENTS:<\/strong>\r\n\r\n \r\n

Events that cannot occur together are said to be mutually exclusive events. Such events do not have any common outcomes.<\/p>\r\n \r\n

Thus, occurrence of one event excludes the occurrence of other event of events.<\/p>\r\n \r\n\r\n\"\"\r\n

Mutually non-exclusive events\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/em><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nINDEPENDENT EVENT:<\/strong>\r\n\r\n \r\n

The occurrence of one event does not change the probability of the occurrence of other event is called independent events<\/p>\r\n \r\n\r\nEvents are independent if\r\n\r\n \r\n\r\nEither P (A\/B) = P (A) or P (B\/A) = P(B)\r\n\r\n \r\n\r\nDEPENDENT EVENT:<\/strong>\r\n\r\n \r\n

It the occurrence of event affects the probability of the occurrence of other event, then the two events are said to be dependent events.<\/p>\r\n \r\n\r\nThe two events will be dependent if either P (A\/B) \u00a0P (A) or P (B\/A) \u00a0P (B)\r\n\r\nAn event may be a simple event<\/em> or a compound event.<\/em> A simple event is also called an elementary<\/em> event<\/em>, and a compound event is also called a composite event<\/em>.\r\n\r\n \r\n\r\nSimple Event<\/strong>\r\n\r\n \r\n

Each of the final outcomes for an experiment is called a simple event.<\/strong> In other words, a simple event includes one and only one outcome. Usually, a simple event is denoted by E<\/em>1<\/em>, E<\/em>2<\/em>, E<\/em>3<\/em> and so forth. However, we can denote it by any of the other letter too, that is, by A, B, C<\/em> and so forth.<\/p>\r\n \r\n\r\nSIMPLE EVENT<\/strong>\r\n\r\n \r\n

An event that includes one and only one of the (final) outcome for an experiment is called a simple event and is usually denoted by E<\/em>1<\/em>.<\/p>\r\nExample describes simple events\r\n\r\n \r\n

Example 2: <\/strong>Suppose we randomly select two persons from the members of a club and observing whether the person selected each time is a man or a woman. Each of the final outcomes (MM, MW,<\/em> WM and WW) <\/em>for this experiment is a simple event. These four events can be denoted by E<\/em>1<\/em>, E<\/em>2<\/em> , E<\/em>3<\/em> and E<\/em>4<\/em> , respectively. Thus,<\/p>\r\n \r\n\r\nE<\/em>1<\/em> = (MM),<\/em>\u00a0E<\/em>2<\/em> = (MW)<\/em>\u00a0 ,\u00a0E<\/em>3<\/em>= (WM),<\/em>\u00a0\u00a0and\u00a0<\/em>\u00a0E<\/em>4<\/em> = (WW)<\/em>\r\n\r\n \r\n\r\nCompound Event<\/strong>\r\n\r\n \r\n\r\nA compound event<\/em><\/strong> consists of more than one outcome.\r\n\r\n \r\n\r\nCOMPOUND EVENT<\/strong>\r\n\r\n \r\n\r\nA Compound event is a collection of more than one outcome for an experiment.\r\n\r\n \r\n\r\nCompound events are denoted by A, B, C, D\u2026\u2026\u2026 or<\/em> by A<\/em>1<\/em>, A<\/em>2<\/em>, A<\/em>3<\/em>, \u2026\u2026.. B<\/em>1<\/em>, B<\/em>2<\/em> B<\/em>3<\/em>\u2026\u2026..<\/em> and so forth.\r\n\r\n \r\n\r\nExamples 3,4 describe compound events.\r\n\r\n \r\n

Example 3: <\/strong>Reconsider Example 2 about selecting two persons from the members of a club and observing whether the person selected each time is a man or a woman. Let <\/span>A<\/em> be the event that at most one man is selected. Event <\/span>A<\/em> will occur if either no man or one man is selected. Hence, The event <\/span>A<\/em> is given by<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nA = {MW, WM, WW}<\/em>\r\n\r\nSince event A<\/em> contains more than one outcome, it is a compound event. The Venn diagram of Figure:\r\n\r\ngives a graphic presentation of compound event A<\/em>.\r\n\r\n\"\"\r\n\r\n \r\n

Example 4: <\/strong>In a group of students some are in favour of statistics and others are against it. Two persons are selected at random from this group and asked whether they are in favour of or against statistics. How many distinct outcomes are possible? Draw a Venn diagram and a tree diagram for the experiment. List all the outcomes included in each of the following events and mention whether they are simple or compound events.<\/p>\r\n \r\n\r\n(a) Both persons are in favor of statistics.\r\n\r\n \r\n\r\n(b) At most one person is against statistics .\r\n\r\n \r\n\r\n(c) Exactly one person is in favor of statistics .\r\n\r\n \r\n\r\nSolution:\u00a0<\/em><\/strong>\u00a0Let:<\/em><\/strong>\r\n\r\n \r\n\r\nF = <\/em>a person is in favor of statistics.\r\n\r\n \r\n\r\nA = <\/em>a person is against statistics.\r\n\r\n \r\n\r\nThis experiment has the following four outcomes.\r\n\r\n \r\n\r\nFF = both persons are in favor of statistics.\r\n\r\n \r\n\r\nFA = the first person is in favor and the second is against.\r\n\r\n \r\n\r\nAF = the first person is against and the second is in favor.\r\n\r\n \r\n\r\nAA = both persons are against statistics.\r\n\r\n \r\n\r\n \r\n\r\nThe Venn and tree diagrams is.<\/strong>\r\n\r\n<\/div>\r\n\"\"\r\n

\r\n\r\n \r\n\r\n(a) The event \u201cboth persons are in favor of statistics\u201d will occur if FF<\/em> is obtained. Thus,\r\n\r\n \r\n\r\nBoth persons are in favor of statistics = {FF}.<\/em>\r\n\r\n \r\n\r\nBecause this event includes only one of the final four outcomes, it is a simple event.\r\n\r\n \r\n

(b) The event \u201cat most one person is against statistics\u201d will occur if either none or one of the person selected is against statistics. Consequently, At most one person is against statistics = {FF, FA, AF}<\/em><\/p>\r\n \r\n\r\nBecause this event include more than one outcome, it is a compound event.\r\n\r\n \r\n

(c) The event \u201cexactly one person is in favour of statistics\u201d will occur if one of the two persons selected is in favour and the other is against statistics. Hence, it includes the following two outcomes.<\/p>\r\n \r\n\r\nExactly one person is in favor\u00a0 of statistics = {FA,\u00a0 AF}<\/em>\r\n\r\n \r\n\r\nBecause this event includes more than one outcome, it is a compound event.\r\n\r\n \r\n\r\nTWO PROPERTIES OF PROBABILITY<\/strong>\r\n\r\n \r\n\r\nThe following are two important properties of probability.\r\n\r\n \r\n\r\n1.\u00a0 \u00a0<\/strong>The probability of an event always lies in the range zero to 1.<\/strong>\r\n\r\n \r\n

Whether it is a simple or a compound event, the probability of an event is never less than zero or greater than 1. Using mathematical notation, we can write this property as follows.<\/p>\r\n\"\"\r\n\r\nAn event that cannot occur has zero probability; such an event is called an impossible event<\/em><\/strong>. An event that is certain to occur has a probability equal to 1 and is called a sure event<\/em><\/strong>. That is,\r\n\r\n<\/div>\r\n \r\n\r\nFor an impossible event <\/span>M<\/em> :\u00a0<\/span>P(M) = <\/em>0<\/span>\r\n\r\nFor a sure event <\/span>C<\/em> :<\/span>P(C) = <\/em>1<\/span>\r\n

\r\n\r\n \r\n\r\n2.\u00a0\u00a0\u00a0 The sum of the probabilities of all simple events (or final outcomes) for a experiment, denoted by <\/strong>P<\/em>(E<\/em>1) is always 1.<\/strong>\r\n\r\n \r\n\r\nThus, for an experiment.\r\n\r\n\"\"\r\n\r\nFrom this property, for the experiment of one toss of\u00a0 a coin.\r\n\r\n \r\n\r\nP(H) + P (T) = 1<\/em>\r\n\r\n \r\n\r\nFor the experiment of two tosses of\u00a0 a coin.\r\n\r\n \r\n\r\nP(HH) + P(HT) + P(TH) + P(TT) = 1<\/em>\r\n\r\n \r\n\r\nFor one game of Tennis by a National Tennis team.<\/span>\r\n\r\n \r\n\r\nP(Win) + P(Loss) + P(Tie) = 1<\/em>\r\n\r\n<\/div>\r\n
\r\n\r\n \r\n\r\nAPPROACHES TO THE DEFINITION OF PROBABILITY<\/strong>\r\n\r\n \r\n\r\n\"\"\r\n\r\nAXIOMATIC APPROACH :<\/strong>\r\n\r\n \r\n
    \r\n \t
  1. The probability of an event always lies in the range zero to 1. The probability of sure event is 1 and that of impossible event is zero i.e. P(S)=1 and P( ) = 0.<\/li>\r\n \t
  2. The sum of the probabilities of all simple events (or final out comes) for an experiment, denoted by \u2211\u00a0\u00a0 (\u00a0\u00a0 ) is always.<\/li>\r\n<\/ol>\r\n \r\n
      \r\n \t
    1. i.e. \u2211\u00a0 (Ei) = P(E1) + P(E2) + \u2026\u2026..+ = 1<\/li>\r\n<\/ol>\r\n \r\n\r\nFor the experiment of two tosses of a coin\r\n\r\n \r\n\r\nP(HH) + P(HT) + P(TH) + P(TT) = 1.\r\n\r\n \r\n\r\nClassical Probability:<\/strong>\r\n\r\n \r\n

      The Classical Probability rule is applied to compute the probabilities of events for an experiment all of whose outcomes are equally likely.<\/p>\r\n \r\n\r\nSome examples illustrate how probabilities of events are calculated using the classical probability rule.\r\n\r\n \r\n\r\n<\/div>\r\n

      <\/div>\r\n
      <\/div>\r\n\"\"\r\n
      \r\n\r\n\"\"\r\n\r\nRelative Frequency Concept of Probability<\/strong>\r\n\r\n \r\n\r\n1.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The probability that the next baby born at a hospital is a girl.\r\n\r\n \r\n\r\n2.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The probability that an 80-year-old person will live for at least one more year.\r\n\r\n \r\n

      3.\u00a0\u00a0\u00a0\u00a0\u00a0 The probability that the tossing of an unbalanced coin will reset in a head. The probability that we will observe a 1- spot if we roll a loaded die.<\/p>\r\n \r\n

      Although the various outcomes for each of these experiments are not equally likely, each of these experiments can be performed again and again to generate data. In such cases, to calculate probabilities we either use past data or generate new data by performing the experiment a large number of time. The relative frequency of an event is used as an approximation for the probability of that event. This method of assigning a probability to an event is called the relative frequency concept<\/em><\/strong> of probability.<\/em><\/strong><\/p>\r\n \r\n

      RELATIVE FREQUENCY AS AN APPROXIMATION OF PROBABILITY<\/strong><\/p>\r\n \r\n

      If an experiment is repeated <\/span>n<\/em> times and an event <\/span>A<\/em> is observed <\/span>f<\/em> times, then, according to the\u00a0<\/span>relative frequency concept of probability : <\/span>P A<\/em>\u00a0 <\/span>n<\/em>f<\/em><\/p>\r\n\r\n<\/div>\r\n

      \r\n\r\n \r\n\r\nSUBJECTIVE PROBABILITY<\/strong>\r\n\r\n \r\n

      Many times we face experiments that neither have equally likely outcomes nor can be repeated to generate data. In such cases, we cannot compute the probabilities of events using the classical probability rule or the relative frequency concept. For example, consider the following probabilities of events.<\/p>\r\n \r\n\r\n1.\u00a0 \u00a0 The probability that John, who is taking statistics this semester, will earn an A<\/em> in this course.\r\n\r\n \r\n\r\n2.\u00a0 \u00a0The probability that Ali will lose the lawsuit the he has filed against his landlord.\r\n\r\n \r\n

      Neither the classical probability rule nor the relative frequency concept of probability can be applied to calculate probabilities for these example. All these examples belong to experiments that have neither equally likely outcomes nor the potential of being repeated. For example, Chirayu, who is taking statistics this semester, will take the test (or tests) only once and based on that he will either earn an A<\/em> or not. The two events \u201che will earn an A\u201d and \u201che will not earn an A\u201d are not equally likely. The probability assigned to an event in such cases is called subjective probability<\/em>.<\/strong> It is based on the individual\u2019s own judgment, experience, information, and belief, Chirayu may assign a high probability to the event that he will earn an A<\/em> in statistics, whereas his instructor may assign a low probability to the same event.<\/p>\r\n \r\n\r\nSUBJECTIVE PROBABILITY<\/strong>\r\n\r\n \r\n

      Subjective probability is the probability assigned to an event based on subjective judgment, experience, information, and belief.<\/p>\r\n \r\n\r\nSubjective probability is assigned arbitrarily. It is usually influenced by the biases, preferences, and experience of the person assigning the probability.\r\n\r\n \r\n\r\nBasic rules of Probability :<\/strong>\r\n\r\n \r\n\r\n1.\u00a0\u00a0 Addition Theorem: When events are mutually exclusive:\r\n\r\n \r\n

      If two events A & A2 are mutually exclusive, the probability of the occurance of either A or A2 is the sum of the individual probability of A1 or A2<\/p>\r\n \r\n\r\ni.e. P(A1UA2) = P(A or A2) + P(A1) + P(A2)\r\n\r\n \r\n\r\nIf we extend it for n mutually exclusive events\r\n\r\n \r\n\r\ni.e. P(A1or A2 or A3 or \u2026\u2026..) = P(A1) + P(A2) + P(A3)\u2026\u2026\r\n\r\n \r\n\r\n2.\u00a0 \u00a0Multiplication Theorem :\r\n\r\n \r\n

      If two events A1 & A2 are independent (i.e. occurrence of one does not affect the other) The probability that both will occur is<\/span><\/p>\r\n \r\n

      P(A1 A2) = P (A1 and A2) = P (A) X P (A2)<\/span><\/p>\r\n\r\n<\/div>\r\n

      \r\n\r\n \r\n\r\nFor n independent events\r\n\r\n \r\n\r\nP(A1\u00a0 A2\u00a0 A3----------) = P(A1 X A2 X A3 --------)\r\n\r\n \r\n\r\n=\u00a0 P(A1) x P(A2) X P(A3) X -------\r\n\r\n \r\n\r\n3.\u00a0 Two complementary events are always mutually exclusive i.e. P(A) + P(\u1fb9) = 1\r\n\r\n \r\n\r\nwhere \u1fb9is complement of A.\r\n\r\n \r\n

      4.\u00a0 \u00a0 \u00a0If n independent events with respective probabilities P1, P2, P3--------Pn are given then probability of at least one of these events happening is<\/p>\r\n \r\n\r\n1-(1-P1). (1-p2).(1-p3) ----------(1-pn).\r\n\r\n \r\n

      5.\u00a0 If events happen in mutually non exclusive way rather not in mutually exclusive way e.g. The probability of A passing in physics is 2\/3 and that of mathematics is 4\/9. If the probability of passing both courses is 2\/5 then the probability of A passing at least one of the courses is<\/p>\r\n \r\n\r\np(M U P)\u00a0 = p(M) + p(P) \u2013 P (M\u00a0 P)\r\n\r\n \r\n\r\n=\u00a0 4\/9 + 2\/3\u00a0 - 2\/5\r\n\r\n \r\n\r\n=\u00a0 32\/45\r\n\r\n \r\n\r\n<\/div>\r\n

      \r\n\r\n\"\"\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nQ.3<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 If card is drawn at random from a standard deck of cards, what is probability that it is a king or a heart.\r\n\r\n \r\n\r\nAns.<\/strong>\u00a0 \u00a0Probability that it is king from deck of card = 524 Probability that it is heart = 524\r\n\r\n \r\n\r\nP (King of heart)\u00a0 = probability that the king of heart is drawn.\r\n\r\n=\u00a0\u00a0\u00a0 521\r\n\r\n \r\n\r\nP (a king or a heart)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = p (king) + p (Heart) \u2013 p (king of heart)\r\n\r\n=\u00a0\u00a0\u00a0 524 + 1352 \u2212 521\r\n\r\n \r\n\r\n=\u00a0\u00a0\u00a0 1652 is the required probability.\r\n\r\n \r\n

      Q.4<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A card is drawn from a standard deck of 52 cards, two suits are observed and card is then replaced before a second one is drawn. What is the probability that both are heart?<\/p>\r\n13\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\r\n\r\nAns.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Probability of each card is heart = 52 = 4\r\n\r\n1\u00a0\u00a0\u00a0\u00a0\u00a0 1\r\n\r\nProbability of 2 hearts drawn is = 4 x 4\r\n\r\n \r\n\r\nQ.5<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 What is the sample space and event in tossing an unbiased coin twice.\r\n\r\n \r\n\r\nAns.<\/strong>\u00a0\u00a0\u00a0\u00a0 Sample space is {HH, HT,\u00a0 TH,\u00a0 TT}\r\n\r\n \r\n\r\nEach possible distinct outcome such as HH, HT,\u00a0 TH & TT is considered as an event.\r\n\r\n \r\n

      Q.6<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Suppose that 80% of all tourists who come to India will visit Delhi. 70% of them will visit Mumbai and 60% of them will visit Both Delhi and Mumbai .What is the probability that he will visit Delhi or Mumbai or both. What is the probability that he will visit neither city?<\/p>\r\n \r\n\r\nAns.<\/strong>\u00a0\u00a0\u00a0\u00a0 Let us define A as the event that the tourist will visit Delhi and B as the event that the tourist will visit Mumbai.\r\n\r\n \r\n\r\nThen\u00a0 \u00a0 P(A) = 0.8,\u00a0 \u00a0P(B) = 0.7\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0\u00a0 P(AB) = 0.6\r\n\r\n \r\n\r\nP (A+B) = probability that the tourist will visit Delhi or Mumbai or both.\r\n\r\n \r\n\r\n=\u00a0 P(A)\u00a0 + P(B)\u00a0 \u2013 P (AB)\r\n\r\n \r\n\r\n=\u00a0 0.8 + 0.7\u00a0 \u2013 0.6\r\n\r\n \r\n\r\n=\u00a0\u00a0\u00a0 0.9\r\n\r\n \r\n\r\nThe probability that he will visit neither city\r\n\r\n \r\n\r\n=\u00a0 1 \u2013 P(A+B)\r\n\r\n \r\n\r\n=\u00a0 1 \u2013 0.9 = 0.1\r\n\r\n \r\n

      Q.7<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A bag contains 8 white and 4 red balls. Five balls are drawn at random. What is the probability that 2 of them are red and 3 white?<\/p>\r\n \r\n\r\nAns.<\/strong>\u00a0\u00a0\u00a0\u00a0 Total no. of balls in bag = 8 + 4 = 12\r\n\r\n \r\n\r\nNumber of balls draws = 5.\r\n\r\n<\/div>\r\n

      \r\n

      5 balls can be drawn from 12 in 12C5 ways, 2 balls can be drawn from 4 red in 4C2 ways, and 3 white balls can be drawn from 8 balls in 8C3 ways<\/p>\r\n \r\n\r\nTherefore, the number of favorable cases are 4C2 x 8C3.\r\n\r\n \r\n\r\nan required probability\r\n\r\n \r\n\r\n= 0.424\r\n\r\n \r\n\r\nQ.8<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A bag contains 2 white and 3 black balls. Four persons A, B, C, D in the order are named. Each draw one ball and do not replace it back. The person to draw a white ball receives Rs. 200. Determine their expectations\r\n\r\n \r\n\r\nAns.<\/strong>\u00a0\u00a0\u00a0\u00a0 Since only 3 black balls are contained in the bag one person must win in the first attempt. Probability that A wins or the ball drawn is white\r\n\r\n=\u00a0\u00a0\u00a0 25.\r\n\r\n \r\n\r\n\u2234\u00a0 \u00a0 A\u2019s expectation = 25 x 200 = Rs. 80\/-\r\n\r\nProbability that A looses and B wins =\u00a0 35 \u00d7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 24 = 103\r\n\r\n\u2234\u00a0 \u00a0 \u00a0B\u2019s expectation = 103\u00a0 \u00a0\u00d7 200 = Rs. 60\/-\r\n\r\n<\/div>\r\n \r\n\r\nProbability that A and B looses but C wins = <\/span>\r\n\r\n3 2 2 1<\/span>\r\n\r\n5435\r\n\r\n \r\n

      \r\n\r\n\u2234\u00a0 \u00a0 \u00a0C\u2019s expectation = 51 \u00d7 200 = Rs. 40\/-\r\n\r\n \r\n\r\nProbability that A, B, and C looses but D wins\r\n\r\n=\u00a0\u00a0\u00a0\u00a0\u00a0 35 \u00d7 24 \u00d7 23 \u00d7 13 \u00d7 22 = 101\r\n\r\n \r\n\r\n\u2234\u00a0 \u00a0 D\u2019s expectation = 101\u00a0 \u00d7 200 = Rs. 20\/-\r\n\r\n \r\n

      Q.9<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A husband and wife appear in an interview for two vacancies in the same post. The probability of husband\u2019s selection is 1\/7 and that of wife\u2019s selection is 1\/5. What is the probability that<\/p>\r\n \r\n\r\n(a)\u00a0 both of them will be selected.\r\n\r\n \r\n\r\n(b) only one of them will be selected, and\r\n\r\n \r\n\r\n(c)\u00a0 None of them will be selected.\r\n\r\n \r\n\r\nAns.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 Let A denote husband\u2019s selection and B denote wife\u2019s selection.\r\n\r\n \r\n\r\n(a) Probability that both of them will be selected. Since events are independent, we have P(A and B) = P(A) = P(B) = 17 \u00d7 15 = 351 = 0.029\r\n\r\n \r\n\r\n(b)\u00a0 Probability that only one of them will be selected.\r\n\r\n=\u00a0\u00a0\u00a0\u00a0\u00a0 p[(A and <\/em>\u0305)] or (B and\u00a0\u00a0 <\/em>\u0305)<\/em>\r\n\r\n\u00a0<\/em>\r\n\r\n=\u00a0\u00a0\u00a0\u00a0\u00a0 p (A and\u00a0 <\/em>\u0305) + P (B and\u00a0\u00a0 <\/em>\u0305)<\/em>\r\n\r\n\u00a0<\/em>\r\n\r\n=\u00a0\u00a0\u00a0\u00a0 P (A) P( <\/em>\u0305) + P(B) P(\u00a0 <\/em>\u0305)<\/em>\r\n\r\n \r\n\r\n= P(A) [1\u2013P(B)] + P(B) [1\u2013P(A)]<\/em>\r\n\r\n \r\n\r\n=\u00a0\u00a0\u00a0\u00a0 17 \u00d7 45 + 15\u00a0 \u00d7 67<\/span>\r\n\r\n \r\n\r\n= 47 \u00d7 45 + 15 \u00d7 67\u00a0 ccc<\/span>\r\n\r\n \r\n\r\n= 354 + 356 = 1035<\/span>\r\n\r\n \r\n\r\n=\u00a0\u00a0 <\/span>0.286<\/em>\r\n\r\n<\/div>\r\n

      \r\n\r\n\u00a0<\/em>\r\n\r\n(c)\u00a0 \u00a0Probability that none of them will be selected\r\n\r\n=\u00a0\u00a0\u00a0\u00a0\u00a0 P(\u00a0 <\/em>\u0305) x P( <\/em>\u0305)<\/em>\r\n\r\n\u00a0<\/em>\r\n\r\n=\u00a0\u00a0\u00a0\u00a0 67 \u00d7 45 = 2435\r\n\r\n\u00a0<\/em>\r\n\r\n=\u00a0\u00a0 0.686<\/em>\r\n\r\n \r\n\r\nMULTIPLICATION RULE :<\/strong>\r\n\r\n \r\n\r\nIf A and B are independent events, the joint probability of A and B occurring is given by :\r\n\r\n \r\n\r\nP(AB) = P(A) x P(B)<\/em>\r\n\r\n \r\n\r\nGENERAL MULTIPLICATION RULE :<\/strong>\r\n\r\n \r\n\r\nP(AB) = P(A) X P (B\/A)<\/em>\r\n\r\n \r\n

      Example : <\/strong>Suppose a company hires both MBAs and non MBAs for the same kind of managerial tasks. After a period of employment some of each category are promoted and some are not. The table gives the proportion of company\u2019s manager\u2019s among the said classes.<\/p>\r\n \r\n\r\n\"\"\r\n

      Calculate P(A\/B) and P(B\/A). Find out whether A and B are independent.<\/em><\/strong><\/p>\r\n \r\n

      Ans. <\/strong>As these probabilities appear in the margin they are called as marginal probabilities. It may observed that marginal probability of A is the probability of the event A and it is the sum of probabilities in which A is involved. It may be observed that<\/p>\r\n \r\n\r\nP (AB) = P(A)\u00a0 x P(B) = 0.7 x 0.60 = 0.42<\/em>\r\n\r\n \r\n\r\nThis is true for all the cells\r\n\r\n \r\n\r\nThus, Probability (MBA and promoted) = P(MBA) x P(promoted)\r\n\r\n \r\n\r\n= 0.7\u00a0 x 0.6\u00a0\u00a0 = 0.42\r\n\r\n<\/div>\r\n \r\n\r\nProbability (MBA and Non Promoted) = P(MBA) x P( Non promoted)\r\n\r\n \r\n\r\n= 0.7\u00a0 x 0.4\u00a0\u00a0 = 0.42\r\n\r\n \r\n\r\nProbability (Non MBA and Non Promoted)= 0.3 x 0.4 = 0.12\r\n\r\n \r\n\r\nWe conclude that the educational qualification and promotional status of managers are independent of each other. We also conclude and calculate.\r\n\r\n \r\n\r\nThus we find P(A\/B) = P(A) and P (B\/A) = P(B), which means A and B are independent.\r\n\r\n \r\n\r\n \r\n\r\nSummary<\/strong>\r\n\r\n \r\n

      In the last ,Probability which measures the likelihood that an event will occur, is an important part of Statitics.It is basis of inferential statistics (from general to specific).Probability Theory is used to evaluate the uncertainity involved in those decisions e.g. estimating next year \u2019s sale for a company is based on many assumptions.Some of which may happen to be true and others may not.So here probability theory and its tool will help to make decision under such conditions of imperfect information and uncertainity.Basic concepts of probability and the rules for computing probability are explained here.<\/p>\r\n ","rendered":"

      \n

       <\/p>\n

      1.\u00a0\u00a0\u00a0\u00a0\u00a0 Introduction<\/p>\n

       <\/p>\n

      2.\u00a0\u00a0\u00a0\u00a0\u00a0 Terminology<\/p>\n

       <\/p>\n

      3.\u00a0\u00a0\u00a0\u00a0\u00a0 Properties of Probability<\/p>\n

       <\/p>\n

      4.\u00a0\u00a0\u00a0\u00a0\u00a0 Approaches To Probability<\/p>\n

       <\/p>\n

      5.\u00a0\u00a0\u00a0\u00a0\u00a0 Problems<\/p>\n

       <\/p>\n

      6.\u00a0\u00a0\u00a0\u00a0\u00a0 Summary<\/p>\n

       <\/p>\n

       <\/p>\n

      Introduction<\/strong><\/p>\n

       <\/p>\n

      Probability which measures the likelihood that an event will occur is an important part of statistics. It is the basis of inferential statistics Here we make decisions under conditions of uncertainty. Probability theory is used to evaluate the uncertainty involved in those decisions. The concept of statistical probability may be characterized by saying that probability do not refer to judgment but predict the possible outcomes of a conceptual experiment, such as tossing a coin or counting the number of telephone calls, estimating next year\u2019s sales for a company is based on many assumptions, some of which may happen to be true and others may not.<\/p>\n

       <\/p>\n

      Probability theory will help us make decisions under such conditions of imperfect information and uncertainty. Combining probability & probability distribution with descriptive statistics will help us make decision about population based on information obtained from samples.<\/p>\n

       <\/p>\n

      TERMINOLOGY:<\/strong><\/p>\n

       <\/p>\n

      RANDOM EXPERIMENT:<\/strong><\/p>\n

       <\/p>\n

      Any experiment performed under normal circumstances which can be repeated under the same conditions and which results in one or only one of many observations. These observations are called outcome of the experiment.<\/span><\/p>\n<\/div>\n

      \n

       <\/p>\n

      OUTCOME:<\/strong><\/p>\n

       <\/p>\n

      Each element of a sample space is called an outcome. It is also called a sample point.<\/p>\n

       <\/p>\n

      SAMPLE SPACE:<\/strong><\/p>\n

       <\/p>\n

      The set of all possible outcomes of a trial (or experiment) is called a sample space and is generally denoted by S.<\/p>\n

       <\/p>\n

      EXPERIMENT, OUTCOME & SAMPLE SPACE:<\/strong><\/p>\n

       <\/p>\n

      Sample space for a weather forecaster to predict the rain is written as<\/p>\n

       <\/p>\n

      S = {rainy day, dry day}<\/em><\/p>\n

       <\/p>\n

      The elements of a sample space are called sample point.<\/p>\n

       <\/p>\n

      Sample space for a student to get success or failure in semester is written as S = {Fail, pass}<\/em><\/p>\n

       <\/p>\n

      Lists of few examples of experiment, outcome & Sample space.<\/p>\n

       <\/p>\n

      Table:<\/p>\n

       <\/p>\n\n\n\n\n\n\n\n\n\n
      Experiment<\/strong><\/td>\nOutcomes<\/strong><\/td>\nSample space<\/strong><\/td>\n<\/tr>\n
      Toss a coin<\/td>\nHead, Tail<\/td>\nS = {Head, Tail}<\/td>\n<\/tr>\n
      Roll a dice<\/td>\n1,2,3,4,5,6<\/td>\nS = {1, \u2026\u2026..6}<\/td>\n<\/tr>\n
      Toss a coin twice<\/td>\nHH, HT, TH, TT<\/td>\nS = {HH, HT, TH, TT}<\/td>\n<\/tr>\n
      Birth of a baby<\/td>\nBoy, Girl<\/td>\nS = {Boy, Girl}<\/td>\n<\/tr>\n
      Take a test<\/td>\nPass, Fail<\/td>\nS = {Pass, Fail}<\/td>\n<\/tr>\n
      Select a student<\/td>\nMale, Female<\/td>\nS = Male, Female<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      The sample space for an experiment can be understood by drawing a venn diagram as well as a tree diagram let us take examples.<\/p>\n

       <\/p>\n

      TOSSING A COIN ONCE:<\/strong><\/p>\n<\/div>\n

      \n

      \"\"\"\"<\/p>\n

       <\/p>\n

       <\/p>\n

       <\/p>\n

      Therefore, the sample space for two tosses of the coin is<\/p>\n

       <\/p>\n

      S = {HH, HT,\u00a0 TH,\u00a0 TT}<\/p>\n

       <\/p>\n

      Example 1 : Suppose we randomly select two babies from a hospital and observe whether a baby selected each time is a baby boy or a baby girl. Write all outcomes for this experiment.<\/p>\n

      \"\"<\/p>\n

      EQUALLY LIKELY EVENTS :<\/strong><\/p>\n

       <\/p>\n

      The event are said to be equally likely when one event does not occur more often than the other.<\/p>\n

       <\/p>\n

      EVENT<\/strong><\/p>\n

       <\/p>\n

      An event is a collection of one or more of the outcomes of an experiment.<\/p>\n

       <\/p>\n

      EVENT:<\/strong><\/p>\n

       <\/p>\n

      An event is a collection of one or more of the outcomes of an experiment. E.g. tossing of a coin is trial and getting Head (H) or Tail (T) is an event.<\/p>\n

       <\/p>\n

      FAVOURABLE CASES:<\/strong><\/p>\n

       <\/p>\n

      The number of cases favorable to an event in any trial is the number of outcomes which entail the happening of the event.<\/p>\n

       <\/p>\n

      EXHAUSTIVE EVENTS:<\/strong><\/p>\n

       <\/p>\n

      The total number of possible outcomes in any trial is called exhaustive events if taken together covers the sample space.<\/p>\n

       <\/p>\n

      MUTUALLY EXCLUSIVE EVENTS:<\/strong><\/p>\n

       <\/p>\n

      Events that cannot occur together are said to be mutually exclusive events. Such events do not have any common outcomes.<\/p>\n

       <\/p>\n

      Thus, occurrence of one event excludes the occurrence of other event of events.<\/p>\n

       <\/p>\n

      \"\"<\/p>\n

      Mutually non-exclusive events\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/em><\/p>\n<\/div>\n

      \n

       <\/p>\n

      INDEPENDENT EVENT:<\/strong><\/p>\n

       <\/p>\n

      The occurrence of one event does not change the probability of the occurrence of other event is called independent events<\/p>\n

       <\/p>\n

      Events are independent if<\/p>\n

       <\/p>\n

      Either P (A\/B) = P (A) or P (B\/A) = P(B)<\/p>\n

       <\/p>\n

      DEPENDENT EVENT:<\/strong><\/p>\n

       <\/p>\n

      It the occurrence of event affects the probability of the occurrence of other event, then the two events are said to be dependent events.<\/p>\n

       <\/p>\n

      The two events will be dependent if either P (A\/B) \u00a0P (A) or P (B\/A) \u00a0P (B)<\/p>\n

      An event may be a simple event<\/em> or a compound event.<\/em> A simple event is also called an elementary<\/em> event<\/em>, and a compound event is also called a composite event<\/em>.<\/p>\n

       <\/p>\n

      Simple Event<\/strong><\/p>\n

       <\/p>\n

      Each of the final outcomes for an experiment is called a simple event.<\/strong> In other words, a simple event includes one and only one outcome. Usually, a simple event is denoted by E<\/em>1<\/em>, E<\/em>2<\/em>, E<\/em>3<\/em> and so forth. However, we can denote it by any of the other letter too, that is, by A, B, C<\/em> and so forth.<\/p>\n

       <\/p>\n

      SIMPLE EVENT<\/strong><\/p>\n

       <\/p>\n

      An event that includes one and only one of the (final) outcome for an experiment is called a simple event and is usually denoted by E<\/em>1<\/em>.<\/p>\n

      Example describes simple events<\/p>\n

       <\/p>\n

      Example 2: <\/strong>Suppose we randomly select two persons from the members of a club and observing whether the person selected each time is a man or a woman. Each of the final outcomes (MM, MW,<\/em> WM and WW) <\/em>for this experiment is a simple event. These four events can be denoted by E<\/em>1<\/em>, E<\/em>2<\/em> , E<\/em>3<\/em> and E<\/em>4<\/em> , respectively. Thus,<\/p>\n

       <\/p>\n

      E<\/em>1<\/em> = (MM),<\/em>\u00a0E<\/em>2<\/em> = (MW)<\/em>\u00a0 ,\u00a0E<\/em>3<\/em>= (WM),<\/em>\u00a0\u00a0and\u00a0<\/em>\u00a0E<\/em>4<\/em> = (WW)<\/em><\/p>\n

       <\/p>\n

      Compound Event<\/strong><\/p>\n

       <\/p>\n

      A compound event<\/em><\/strong> consists of more than one outcome.<\/p>\n

       <\/p>\n

      COMPOUND EVENT<\/strong><\/p>\n

       <\/p>\n

      A Compound event is a collection of more than one outcome for an experiment.<\/p>\n

       <\/p>\n

      Compound events are denoted by A, B, C, D\u2026\u2026\u2026 or<\/em> by A<\/em>1<\/em>, A<\/em>2<\/em>, A<\/em>3<\/em>, \u2026\u2026.. B<\/em>1<\/em>, B<\/em>2<\/em> B<\/em>3<\/em>\u2026\u2026..<\/em> and so forth.<\/p>\n

       <\/p>\n

      Examples 3,4 describe compound events.<\/p>\n

       <\/p>\n

      Example 3: <\/strong>Reconsider Example 2 about selecting two persons from the members of a club and observing whether the person selected each time is a man or a woman. Let <\/span>A<\/em> be the event that at most one man is selected. Event <\/span>A<\/em> will occur if either no man or one man is selected. Hence, The event <\/span>A<\/em> is given by<\/span><\/p>\n<\/div>\n

      \n

       <\/p>\n

      A = {MW, WM, WW}<\/em><\/p>\n

      Since event A<\/em> contains more than one outcome, it is a compound event. The Venn diagram of Figure:<\/p>\n

      gives a graphic presentation of compound event A<\/em>.<\/p>\n

      \"\"<\/p>\n

       <\/p>\n

      Example 4: <\/strong>In a group of students some are in favour of statistics and others are against it. Two persons are selected at random from this group and asked whether they are in favour of or against statistics. How many distinct outcomes are possible? Draw a Venn diagram and a tree diagram for the experiment. List all the outcomes included in each of the following events and mention whether they are simple or compound events.<\/p>\n

       <\/p>\n

      (a) Both persons are in favor of statistics.<\/p>\n

       <\/p>\n

      (b) At most one person is against statistics .<\/p>\n

       <\/p>\n

      (c) Exactly one person is in favor of statistics .<\/p>\n

       <\/p>\n

      Solution:\u00a0<\/em><\/strong>\u00a0Let:<\/em><\/strong><\/p>\n

       <\/p>\n

      F = <\/em>a person is in favor of statistics.<\/p>\n

       <\/p>\n

      A = <\/em>a person is against statistics.<\/p>\n

       <\/p>\n

      This experiment has the following four outcomes.<\/p>\n

       <\/p>\n

      FF = both persons are in favor of statistics.<\/p>\n

       <\/p>\n

      FA = the first person is in favor and the second is against.<\/p>\n

       <\/p>\n

      AF = the first person is against and the second is in favor.<\/p>\n

       <\/p>\n

      AA = both persons are against statistics.<\/p>\n

       <\/p>\n

       <\/p>\n

      The Venn and tree diagrams is.<\/strong><\/p>\n<\/div>\n

      \"\"<\/p>\n

      \n

       <\/p>\n

      (a) The event \u201cboth persons are in favor of statistics\u201d will occur if FF<\/em> is obtained. Thus,<\/p>\n

       <\/p>\n

      Both persons are in favor of statistics = {FF}.<\/em><\/p>\n

       <\/p>\n

      Because this event includes only one of the final four outcomes, it is a simple event.<\/p>\n

       <\/p>\n

      (b) The event \u201cat most one person is against statistics\u201d will occur if either none or one of the person selected is against statistics. Consequently, At most one person is against statistics = {FF, FA, AF}<\/em><\/p>\n

       <\/p>\n

      Because this event include more than one outcome, it is a compound event.<\/p>\n

       <\/p>\n

      (c) The event \u201cexactly one person is in favour of statistics\u201d will occur if one of the two persons selected is in favour and the other is against statistics. Hence, it includes the following two outcomes.<\/p>\n

       <\/p>\n

      Exactly one person is in favor\u00a0 of statistics = {FA,\u00a0 AF}<\/em><\/p>\n

       <\/p>\n

      Because this event includes more than one outcome, it is a compound event.<\/p>\n

       <\/p>\n

      TWO PROPERTIES OF PROBABILITY<\/strong><\/p>\n

       <\/p>\n

      The following are two important properties of probability.<\/p>\n

       <\/p>\n

      1.\u00a0 \u00a0<\/strong>The probability of an event always lies in the range zero to 1.<\/strong><\/p>\n

       <\/p>\n

      Whether it is a simple or a compound event, the probability of an event is never less than zero or greater than 1. Using mathematical notation, we can write this property as follows.<\/p>\n

      \"\"<\/p>\n

      An event that cannot occur has zero probability; such an event is called an impossible event<\/em><\/strong>. An event that is certain to occur has a probability equal to 1 and is called a sure event<\/em><\/strong>. That is,<\/p>\n<\/div>\n

       <\/p>\n

      For an impossible event <\/span>M<\/em> :\u00a0<\/span>P(M) = <\/em>0<\/span><\/p>\n

      For a sure event <\/span>C<\/em> :<\/span>P(C) = <\/em>1<\/span><\/p>\n

      \n

       <\/p>\n

      2.\u00a0\u00a0\u00a0 The sum of the probabilities of all simple events (or final outcomes) for a experiment, denoted by <\/strong>P<\/em>(E<\/em>1) is always 1.<\/strong><\/p>\n

       <\/p>\n

      Thus, for an experiment.<\/p>\n

      \"\"<\/p>\n

      From this property, for the experiment of one toss of\u00a0 a coin.<\/p>\n

       <\/p>\n

      P(H) + P (T) = 1<\/em><\/p>\n

       <\/p>\n

      For the experiment of two tosses of\u00a0 a coin.<\/p>\n

       <\/p>\n

      P(HH) + P(HT) + P(TH) + P(TT) = 1<\/em><\/p>\n

       <\/p>\n

      For one game of Tennis by a National Tennis team.<\/span><\/p>\n

       <\/p>\n

      P(Win) + P(Loss) + P(Tie) = 1<\/em><\/p>\n<\/div>\n

      \n

       <\/p>\n

      APPROACHES TO THE DEFINITION OF PROBABILITY<\/strong><\/p>\n

       <\/p>\n

      \"\"<\/p>\n

      AXIOMATIC APPROACH :<\/strong><\/p>\n

       <\/p>\n

        \n
      1. The probability of an event always lies in the range zero to 1. The probability of sure event is 1 and that of impossible event is zero i.e. P(S)=1 and P( ) = 0.<\/li>\n
      2. The sum of the probabilities of all simple events (or final out comes) for an experiment, denoted by \u2211\u00a0\u00a0 (\u00a0\u00a0 ) is always.<\/li>\n<\/ol>\n

         <\/p>\n

          \n
        1. i.e. \u2211\u00a0 (Ei) = P(E1) + P(E2) + \u2026\u2026..+ = 1<\/li>\n<\/ol>\n

           <\/p>\n

          For the experiment of two tosses of a coin<\/p>\n

           <\/p>\n

          P(HH) + P(HT) + P(TH) + P(TT) = 1.<\/p>\n

           <\/p>\n

          Classical Probability:<\/strong><\/p>\n

           <\/p>\n

          The Classical Probability rule is applied to compute the probabilities of events for an experiment all of whose outcomes are equally likely.<\/p>\n

           <\/p>\n

          Some examples illustrate how probabilities of events are calculated using the classical probability rule.<\/p>\n

           <\/p>\n<\/div>\n

          <\/div>\n
          <\/div>\n

          \"\"<\/p>\n

          \n

          \"\"<\/p>\n

          Relative Frequency Concept of Probability<\/strong><\/p>\n

           <\/p>\n

          1.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The probability that the next baby born at a hospital is a girl.<\/p>\n

           <\/p>\n

          2.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The probability that an 80-year-old person will live for at least one more year.<\/p>\n

           <\/p>\n

          3.\u00a0\u00a0\u00a0\u00a0\u00a0 The probability that the tossing of an unbalanced coin will reset in a head. The probability that we will observe a 1- spot if we roll a loaded die.<\/p>\n

           <\/p>\n

          Although the various outcomes for each of these experiments are not equally likely, each of these experiments can be performed again and again to generate data. In such cases, to calculate probabilities we either use past data or generate new data by performing the experiment a large number of time. The relative frequency of an event is used as an approximation for the probability of that event. This method of assigning a probability to an event is called the relative frequency concept<\/em><\/strong> of probability.<\/em><\/strong><\/p>\n

           <\/p>\n

          RELATIVE FREQUENCY AS AN APPROXIMATION OF PROBABILITY<\/strong><\/p>\n

           <\/p>\n

          If an experiment is repeated <\/span>n<\/em> times and an event <\/span>A<\/em> is observed <\/span>f<\/em> times, then, according to the\u00a0<\/span>relative frequency concept of probability : <\/span>P A<\/em>\u00a0 <\/span>n<\/em>f<\/em><\/p>\n<\/div>\n

          \n

           <\/p>\n

          SUBJECTIVE PROBABILITY<\/strong><\/p>\n

           <\/p>\n

          Many times we face experiments that neither have equally likely outcomes nor can be repeated to generate data. In such cases, we cannot compute the probabilities of events using the classical probability rule or the relative frequency concept. For example, consider the following probabilities of events.<\/p>\n

           <\/p>\n

          1.\u00a0 \u00a0 The probability that John, who is taking statistics this semester, will earn an A<\/em> in this course.<\/p>\n

           <\/p>\n

          2.\u00a0 \u00a0The probability that Ali will lose the lawsuit the he has filed against his landlord.<\/p>\n

           <\/p>\n

          Neither the classical probability rule nor the relative frequency concept of probability can be applied to calculate probabilities for these example. All these examples belong to experiments that have neither equally likely outcomes nor the potential of being repeated. For example, Chirayu, who is taking statistics this semester, will take the test (or tests) only once and based on that he will either earn an A<\/em> or not. The two events \u201che will earn an A\u201d and \u201che will not earn an A\u201d are not equally likely. The probability assigned to an event in such cases is called subjective probability<\/em>.<\/strong> It is based on the individual\u2019s own judgment, experience, information, and belief, Chirayu may assign a high probability to the event that he will earn an A<\/em> in statistics, whereas his instructor may assign a low probability to the same event.<\/p>\n

           <\/p>\n

          SUBJECTIVE PROBABILITY<\/strong><\/p>\n

           <\/p>\n

          Subjective probability is the probability assigned to an event based on subjective judgment, experience, information, and belief.<\/p>\n

           <\/p>\n

          Subjective probability is assigned arbitrarily. It is usually influenced by the biases, preferences, and experience of the person assigning the probability.<\/p>\n

           <\/p>\n

          Basic rules of Probability :<\/strong><\/p>\n

           <\/p>\n

          1.\u00a0\u00a0 Addition Theorem: When events are mutually exclusive:<\/p>\n

           <\/p>\n

          If two events A & A2 are mutually exclusive, the probability of the occurance of either A or A2 is the sum of the individual probability of A1 or A2<\/p>\n

           <\/p>\n

          i.e. P(A1UA2) = P(A or A2) + P(A1) + P(A2)<\/p>\n

           <\/p>\n

          If we extend it for n mutually exclusive events<\/p>\n

           <\/p>\n

          i.e. P(A1or A2 or A3 or \u2026\u2026..) = P(A1) + P(A2) + P(A3)\u2026\u2026<\/p>\n

           <\/p>\n

          2.\u00a0 \u00a0Multiplication Theorem :<\/p>\n

           <\/p>\n

          If two events A1 & A2 are independent (i.e. occurrence of one does not affect the other) The probability that both will occur is<\/span><\/p>\n

           <\/p>\n

          P(A1 A2) = P (A1 and A2) = P (A) X P (A2)<\/span><\/p>\n<\/div>\n

          \n

           <\/p>\n

          For n independent events<\/p>\n

           <\/p>\n

          P(A1\u00a0 A2\u00a0 A3———-) = P(A1 X A2 X A3 ——–)<\/p>\n

           <\/p>\n

          =\u00a0 P(A1) x P(A2) X P(A3) X ——-<\/p>\n

           <\/p>\n

          3.\u00a0 Two complementary events are always mutually exclusive i.e. P(A) + P(\u1fb9) = 1<\/p>\n

           <\/p>\n

          where \u1fb9is complement of A.<\/p>\n

           <\/p>\n

          4.\u00a0 \u00a0 \u00a0If n independent events with respective probabilities P1, P2, P3——–Pn are given then probability of at least one of these events happening is<\/p>\n

           <\/p>\n

          1-(1-P1). (1-p2).(1-p3) ———-(1-pn).<\/p>\n

           <\/p>\n

          5.\u00a0 If events happen in mutually non exclusive way rather not in mutually exclusive way e.g. The probability of A passing in physics is 2\/3 and that of mathematics is 4\/9. If the probability of passing both courses is 2\/5 then the probability of A passing at least one of the courses is<\/p>\n

           <\/p>\n

          p(M U P)\u00a0 = p(M) + p(P) \u2013 P (M\u00a0 P)<\/p>\n

           <\/p>\n

          =\u00a0 4\/9 + 2\/3\u00a0 – 2\/5<\/p>\n

           <\/p>\n

          =\u00a0 32\/45<\/p>\n

           <\/p>\n<\/div>\n

          \n

          \"\"<\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

           <\/p>\n

          Q.3<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 If card is drawn at random from a standard deck of cards, what is probability that it is a king or a heart.<\/p>\n

           <\/p>\n

          Ans.<\/strong>\u00a0 \u00a0Probability that it is king from deck of card = 524 Probability that it is heart = 524<\/p>\n

           <\/p>\n

          P (King of heart)\u00a0 = probability that the king of heart is drawn.<\/p>\n

          =\u00a0\u00a0\u00a0 521<\/p>\n

           <\/p>\n

          P (a king or a heart)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = p (king) + p (Heart) \u2013 p (king of heart)<\/p>\n

          =\u00a0\u00a0\u00a0 524 + 1352 \u2212 521<\/p>\n

           <\/p>\n

          =\u00a0\u00a0\u00a0 1652 is the required probability.<\/p>\n

           <\/p>\n

          Q.4<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A card is drawn from a standard deck of 52 cards, two suits are observed and card is then replaced before a second one is drawn. What is the probability that both are heart?<\/p>\n

          13\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1<\/p>\n

          Ans.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Probability of each card is heart = 52 = 4<\/p>\n

          1\u00a0\u00a0\u00a0\u00a0\u00a0 1<\/p>\n

          Probability of 2 hearts drawn is = 4 x 4<\/p>\n

           <\/p>\n

          Q.5<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 What is the sample space and event in tossing an unbiased coin twice.<\/p>\n

           <\/p>\n

          Ans.<\/strong>\u00a0\u00a0\u00a0\u00a0 Sample space is {HH, HT,\u00a0 TH,\u00a0 TT}<\/p>\n

           <\/p>\n

          Each possible distinct outcome such as HH, HT,\u00a0 TH & TT is considered as an event.<\/p>\n

           <\/p>\n

          Q.6<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Suppose that 80% of all tourists who come to India will visit Delhi. 70% of them will visit Mumbai and 60% of them will visit Both Delhi and Mumbai .What is the probability that he will visit Delhi or Mumbai or both. What is the probability that he will visit neither city?<\/p>\n

           <\/p>\n

          Ans.<\/strong>\u00a0\u00a0\u00a0\u00a0 Let us define A as the event that the tourist will visit Delhi and B as the event that the tourist will visit Mumbai.<\/p>\n

           <\/p>\n

          Then\u00a0 \u00a0 P(A) = 0.8,\u00a0 \u00a0P(B) = 0.7\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0\u00a0 P(AB) = 0.6<\/p>\n

           <\/p>\n

          P (A+B) = probability that the tourist will visit Delhi or Mumbai or both.<\/p>\n

           <\/p>\n

          =\u00a0 P(A)\u00a0 + P(B)\u00a0 \u2013 P (AB)<\/p>\n

           <\/p>\n

          =\u00a0 0.8 + 0.7\u00a0 \u2013 0.6<\/p>\n

           <\/p>\n

          =\u00a0\u00a0\u00a0 0.9<\/p>\n

           <\/p>\n

          The probability that he will visit neither city<\/p>\n

           <\/p>\n

          =\u00a0 1 \u2013 P(A+B)<\/p>\n

           <\/p>\n

          =\u00a0 1 \u2013 0.9 = 0.1<\/p>\n

           <\/p>\n

          Q.7<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A bag contains 8 white and 4 red balls. Five balls are drawn at random. What is the probability that 2 of them are red and 3 white?<\/p>\n

           <\/p>\n

          Ans.<\/strong>\u00a0\u00a0\u00a0\u00a0 Total no. of balls in bag = 8 + 4 = 12<\/p>\n

           <\/p>\n

          Number of balls draws = 5.<\/p>\n<\/div>\n

          \n

          5 balls can be drawn from 12 in 12C5 ways, 2 balls can be drawn from 4 red in 4C2 ways, and 3 white balls can be drawn from 8 balls in 8C3 ways<\/p>\n

           <\/p>\n

          Therefore, the number of favorable cases are 4C2 x 8C3.<\/p>\n

           <\/p>\n

          an required probability<\/p>\n

           <\/p>\n

          = 0.424<\/p>\n

           <\/p>\n

          Q.8<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A bag contains 2 white and 3 black balls. Four persons A, B, C, D in the order are named. Each draw one ball and do not replace it back. The person to draw a white ball receives Rs. 200. Determine their expectations<\/p>\n

           <\/p>\n

          Ans.<\/strong>\u00a0\u00a0\u00a0\u00a0 Since only 3 black balls are contained in the bag one person must win in the first attempt. Probability that A wins or the ball drawn is white<\/p>\n

          =\u00a0\u00a0\u00a0 25.<\/p>\n

           <\/p>\n

          \u2234\u00a0 \u00a0 A\u2019s expectation = 25 x 200 = Rs. 80\/-<\/p>\n

          Probability that A looses and B wins =\u00a0 35 \u00d7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 24 = 103<\/p>\n

          \u2234\u00a0 \u00a0 \u00a0B\u2019s expectation = 103\u00a0 \u00a0\u00d7 200 = Rs. 60\/-<\/p>\n<\/div>\n

           <\/p>\n

          Probability that A and B looses but C wins = <\/span><\/p>\n

          3 2 2 1<\/span><\/p>\n

          5435<\/p>\n

           <\/p>\n

          \n

          \u2234\u00a0 \u00a0 \u00a0C\u2019s expectation = 51 \u00d7 200 = Rs. 40\/-<\/p>\n

           <\/p>\n

          Probability that A, B, and C looses but D wins<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0\u00a0 35 \u00d7 24 \u00d7 23 \u00d7 13 \u00d7 22 = 101<\/p>\n

           <\/p>\n

          \u2234\u00a0 \u00a0 D\u2019s expectation = 101\u00a0 \u00d7 200 = Rs. 20\/-<\/p>\n

           <\/p>\n

          Q.9<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A husband and wife appear in an interview for two vacancies in the same post. The probability of husband\u2019s selection is 1\/7 and that of wife\u2019s selection is 1\/5. What is the probability that<\/p>\n

           <\/p>\n

          (a)\u00a0 both of them will be selected.<\/p>\n

           <\/p>\n

          (b) only one of them will be selected, and<\/p>\n

           <\/p>\n

          (c)\u00a0 None of them will be selected.<\/p>\n

           <\/p>\n

          Ans.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 Let A denote husband\u2019s selection and B denote wife\u2019s selection.<\/p>\n

           <\/p>\n

          (a) Probability that both of them will be selected. Since events are independent, we have P(A and B) = P(A) = P(B) = 17 \u00d7 15 = 351 = 0.029<\/p>\n

           <\/p>\n

          (b)\u00a0 Probability that only one of them will be selected.<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0\u00a0 p[(A and <\/em>\u0305)] or (B and\u00a0\u00a0 <\/em>\u0305)<\/em><\/p>\n

          \u00a0<\/em><\/p>\n

          =\u00a0\u00a0\u00a0\u00a0\u00a0 p (A and\u00a0 <\/em>\u0305) + P (B and\u00a0\u00a0 <\/em>\u0305)<\/em><\/p>\n

          \u00a0<\/em><\/p>\n

          =\u00a0\u00a0\u00a0\u00a0 P (A) P( <\/em>\u0305) + P(B) P(\u00a0 <\/em>\u0305)<\/em><\/p>\n

           <\/p>\n

          = P(A) [1\u2013P(B)] + P(B) [1\u2013P(A)]<\/em><\/p>\n

           <\/p>\n

          =\u00a0\u00a0\u00a0\u00a0 17 \u00d7 45 + 15\u00a0 \u00d7 67<\/span><\/p>\n

           <\/p>\n

          = 47 \u00d7 45 + 15 \u00d7 67\u00a0 ccc<\/span><\/p>\n

           <\/p>\n

          = 354 + 356 = 1035<\/span><\/p>\n

           <\/p>\n

          =\u00a0\u00a0 <\/span>0.286<\/em><\/p>\n<\/div>\n

          \n

          \u00a0<\/em><\/p>\n

          (c)\u00a0 \u00a0Probability that none of them will be selected<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0\u00a0 P(\u00a0 <\/em>\u0305) x P( <\/em>\u0305)<\/em><\/p>\n

          \u00a0<\/em><\/p>\n

          =\u00a0\u00a0\u00a0\u00a0 67 \u00d7 45 = 2435<\/p>\n

          \u00a0<\/em><\/p>\n

          =\u00a0\u00a0 0.686<\/em><\/p>\n

           <\/p>\n

          MULTIPLICATION RULE :<\/strong><\/p>\n

           <\/p>\n

          If A and B are independent events, the joint probability of A and B occurring is given by :<\/p>\n

           <\/p>\n

          P(AB) = P(A) x P(B)<\/em><\/p>\n

           <\/p>\n

          GENERAL MULTIPLICATION RULE :<\/strong><\/p>\n

           <\/p>\n

          P(AB) = P(A) X P (B\/A)<\/em><\/p>\n

           <\/p>\n

          Example : <\/strong>Suppose a company hires both MBAs and non MBAs for the same kind of managerial tasks. After a period of employment some of each category are promoted and some are not. The table gives the proportion of company\u2019s manager\u2019s among the said classes.<\/p>\n

           <\/p>\n

          \"\"<\/p>\n

          Calculate P(A\/B) and P(B\/A). Find out whether A and B are independent.<\/em><\/strong><\/p>\n

           <\/p>\n

          Ans. <\/strong>As these probabilities appear in the margin they are called as marginal probabilities. It may observed that marginal probability of A is the probability of the event A and it is the sum of probabilities in which A is involved. It may be observed that<\/p>\n

           <\/p>\n

          P (AB) = P(A)\u00a0 x P(B) = 0.7 x 0.60 = 0.42<\/em><\/p>\n

           <\/p>\n

          This is true for all the cells<\/p>\n

           <\/p>\n

          Thus, Probability (MBA and promoted) = P(MBA) x P(promoted)<\/p>\n

           <\/p>\n

          = 0.7\u00a0 x 0.6\u00a0\u00a0 = 0.42<\/p>\n<\/div>\n

           <\/p>\n

          Probability (MBA and Non Promoted) = P(MBA) x P( Non promoted)<\/p>\n

           <\/p>\n

          = 0.7\u00a0 x 0.4\u00a0\u00a0 = 0.42<\/p>\n

           <\/p>\n

          Probability (Non MBA and Non Promoted)= 0.3 x 0.4 = 0.12<\/p>\n

           <\/p>\n

          We conclude that the educational qualification and promotional status of managers are independent of each other. We also conclude and calculate.<\/p>\n

           <\/p>\n

          Thus we find P(A\/B) = P(A) and P (B\/A) = P(B), which means A and B are independent.<\/p>\n

           <\/p>\n

           <\/p>\n

          Summary<\/strong><\/p>\n

           <\/p>\n

          In the last ,Probability which measures the likelihood that an event will occur, is an important part of Statitics.It is basis of inferential statistics (from general to specific).Probability Theory is used to evaluate the uncertainity involved in those decisions e.g. estimating next year \u2019s sale for a company is based on many assumptions.Some of which may happen to be true and others may not.So here probability theory and its tool will help to make decision under such conditions of imperfect information and uncertainity.Basic concepts of probability and the rules for computing probability are explained here.<\/p>\n

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