{"id":123,"date":"2018-10-30T06:40:18","date_gmt":"2018-10-30T06:40:18","guid":{"rendered":"http:\/\/mgmtp15.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=123"},"modified":"2018-10-30T07:17:11","modified_gmt":"2018-10-30T07:17:11","slug":"discrete-probability-distribution-binomial-distribution","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/chapter\/discrete-probability-distribution-binomial-distribution\/","title":{"rendered":"Discrete Probability Distribution : Binomial Distribution"},"content":{"raw":"
\r\n\r\n \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\u00a0\u00a0 Binomial Probability Distribution\r\n\r\n \r\n\r\n3.\u00a0\u00a0\u00a0\u00a0\u00a0 Binomial Experiment\r\n\r\n \r\n\r\n4.\u00a0\u00a0\u00a0\u00a0\u00a0 Conditions of Binomial Experiment\r\n\r\n \r\n\r\n5.\u00a0\u00a0\u00a0\u00a0\u00a0 Related Problems\r\n\r\n \r\n\r\n6.\u00a0\u00a0\u00a0\u00a0\u00a0 Summary\r\n\r\n \r\n\r\nIntroduction:<\/strong>\r\n\r\n \r\n\r\nTHE BINOMIAL PROBABILITY DISTRIBUTION<\/strong>\r\n\r\n \r\n

The binomial probability distribution is one of the most widely used discrete probability distributions. It is applied to find the probability that an outcome will occur x times in n performances of an experiment. For example, given that the probability is 0.5 that a mobile manufactured at a firm is defective, we may be interested in finding the probability that in a random sample of three mobiles manufactured at this firm, exactly one will be defective. As a second example, we may be interested in finding the probability that a baseball player, with a batting average of 0.250, will have no hits in 10 trips to the plate.<\/p>\r\n \r\n

To apply the binomial probability distribution, the random variable x must be a discrete dichotomous random variable. In other words, the variable must be a discrete random variable and each repetition of the experiment must result in one of two possible\u00a0outcomes. The binomial distribution is applied to experiments that satisfy the four conditions of a binomial experiment. (These conditions are described here.) Each repetition of a binomial experiment is called a trial or a Bernoulli trial (after Jacob Bernoulli). For, examples, if an experiment is defined as one toss of a coin and this experiment is repeated 10 times, then each repetition ( toss) is called a trial. Consequently, there are 10 total trials for this experiment.<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nTHE BINOMIAL EXPERIMENT<\/strong>\r\n\r\n \r\n\r\nAn experiment that satisfies the following four conditions is called a binomial experiment.\r\n\r\n \r\n

1.There are n identical trials, In other words, the given experiment is repeated n times. All these repetitions<\/p>\r\nare performed under identical conditions.\r\n

2. Each trial has two and only two outcomes. These outcomes are usually called a success and failure.<\/p>\r\n3. The probability of success is denoted by p and that of failure by q, and p + q = 1, The probability p and q\r\n

remain constant for each trial.<\/p>\r\n

4. The trials are independent. In other words, the outcome of one trial does not affect the outcome of<\/p>\r\nanother trial.\r\n\r\n \r\n\r\nCONDITION OF A BINOMIAL EXPERIMENT<\/strong>\r\n\r\n \r\n\r\nA BINOMIAL EXPPERIMENT MUST SATISFY THE FOOLLOWING FOUR CONDITIONS<\/em><\/strong>.\r\n\r\n \r\n\r\n1.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 There are n identical trials.\r\n\r\n \r\n\r\n2.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Each trial has only two possible outcomes.\r\n\r\n \r\n\r\n3.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The probabilities of the two outcomes remain constant.\r\n\r\n \r\n\r\n4.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The trials are independent.\r\n\r\n \r\n

Note that one of the two outcomes of a trial is called a success and the other a failure. Notice that a success does not mean that the corresponding outcome is considered favorable or desirable. Similarly, a failure does not necessarily refer to an unfavorable or undesirable outcome. Success and failure are simply the names used to denote the two possible outcomes of a trial. The outcome to which the question refers is usually called a success; the outcome to which it does not refer is called a failure.<\/p>\r\n \r\n\r\nEXAMPLE 1: Consider the experiment consisting of 10 tosses of a coin. Determine if it is a binomial experiment.\r\n\r\n \r\n\r\nSolution; As described below, the experiment consisting of 10 tosses of a coin satisfies all four conditions of a binomial experiment.\r\n\r\n \r\n

    \r\n \t
  1. There are a total of 10 trials (tosses), and they are all identical. All 10 tosses are performed under identical conditions.<\/li>\r\n \t
  2. Each trial (toss) has only two possible outcomes: a head and a tail. Let a head be called a success and a tail be called a failure.<\/li>\r\n \t
  3. The probability of obtaining a head (a success) is \u00bd and that of a tail ( is failure) is \u00bd for any toss. That is,<\/li>\r\n<\/ol>\r\n \r\n\r\nP = P(H) = \u00bd and q = P(T) =\u00a0 \u00bd\r\n\r\n \r\n\r\nThe sum of these two probabilities is 1.0. Also, these probabilities remain the same for each toss.<\/span>\r\n\r\n \r\n\r\nThe trials (tosses) are independent. The result of any preceding toss has no bearing on the result of any succeeding toss.<\/span>\r\n\r\n \r\n\r\nConsequently, the experiment consisting of 10 tosses is a binomial experiment.<\/span>\r\n\r\n<\/div>\r\n
    \r\n\r\n \r\n

    EXAMPLE 2: Five percent of all mobiles manufactured by a large electronics company are defective. Three mobiles are randomly selected from the production line of this company. The selected mobiles are inspected to determine if each of them is defective or good. Is this experiment a binomial experiment?<\/p>\r\n \r\n\r\nSolution<\/strong>\r\n\r\n \r\n

      \r\n \t
    1. This example consists of three identical trials. A trial represents the selection of a mobile.<\/li>\r\n \t
    2. Each trial has two outcomes:\u00a0 A mobile is defective or is a mobile good.\u00a0 Let a defective mobile be called a success and a good mobile\u00a0\u00a0\u00a0 be called a failure.<\/li>\r\n \t
    3. Five percent of all mobiles are defective. So, the probability p that is a defective mobile is .05. As a result, the probability q that a mobile is good is .95. These two probabilities add up to 1.<\/li>\r\n \t
    4. Each trial (mobile) is independent. In other words, if one mobile is defective it does not affect the outcome of another mobile being defective or good. This is so because the size of the population is very large as compared to the sample size.<\/li>\r\n<\/ol>\r\nSince all four conditions of a binomial experiment are satisfied, this is an example of a binomial experiment.\r\n\r\n \r\n\r\nTHE BINOMIAL PROBABILITY DISTRIBUTION AND BINOMIAL FORMULA<\/strong>\r\n\r\n \r\n

      The random variable x that represents the number of successes in n trials for a binomial experiment is called a binomial random variable. The probability distribution of x in such experiments is called the binomial probability distribution or simply binomial distribution. Thus, the binomial probability distribution is applied to find the probability of x successes in n trials for a binomial experiment.<\/p>\r\n \r\n

      To solve binomial problem, we determine the values of n, x, n \u2013 x, p and q then substitute these values in the binomial formula. To find the value of (n<\/em>C<\/em>r<\/em>). We can use either the combinations formula or the table of combinations (Table II)<\/p>\r\n \r\n

      To find the probability of x successes in n trials for a binomial experiment, the only values needed are those of n and p. These are called the parameters of the binomial probability distribution or simply the binomial parameters. The value of q is obtained by subtracting the value of p from, 1.0 Thus, q = 1-p.<\/p>\r\n \r\n\r\nBINOMIAL DISTRIBUTION<\/strong>\r\n\r\n \r\n\r\nRepeated Independent Trials with two Outcomes<\/strong>\r\n\r\n \r\n

      Consider an experiment of repeated trials in which each trial has only two<\/em> possible outcome. Such a situation always arises when we deal with testing of items. Each trial indicates\u00a0whether the item is defective or non-defective. We can define the two outcomes as a \u201csuccess\u201d or \u201cfailure\u201d.<\/span><\/p>\r\n\r\n<\/div>\r\n

      \r\n\r\n \r\n

      These trials are independent as the outcome of any trial does not depend upon the outcome of the previous trial just as in tossing of a coin or drawing a card in succession from a pack of card if the card is replaced every time before the next draw. These independent repeated trials are called Bernoulli trials named after swiss mathematician Jacob Bernoulli. Further the probability of success or that of failure remains constant in each trial.<\/p>\r\n \r\n

      Let a trial be repeated so that we have a series if n<\/em> trials. Let p<\/em> be the Probability of success and q<\/em> that of failure in each Bernoulli trial so that p + q = 1.<\/em><\/p>\r\n \r\n

      It is required to find the probability of r<\/em> successes in n<\/em> trials. If S<\/em> denotes the success and F<\/em> the failure, then the probability of r<\/em> successes and (n<\/em> \u2013<\/em> r<\/em>) failures in a given order say.<\/p>\r\n \r\n

      SFSSFS \u2026..<\/span><\/p>\r\nis pqppqp \u2026\u2026 = pr qn\u2013r\r\n

      as the number is p\u2019s<\/em> is equal to the number of successes, that is, r<\/em> and the number of q\u2019s <\/em>is equal to the number of failures, that is, n \u2013 r<\/em>.<\/p>\r\n \r\n

      In order to find the required probability, we have to count all different orders, in which r<\/em> successes and (n<\/em> \u2013<\/em><\/p>\r\n

      r<\/em>) failures can occur. This is in fact all possible arrangements of n<\/em> letters.<\/p>\r\n\r\n<\/div>\r\n

      \r\n\r\nWhich r are S and (n \u2013 r) are F and is equal to !\/ !( \u2212 )! i.e.\r\n\r\nHence, by addition theorem, the probability of getting r<\/em> successes in a trial is\r\n\r\n \r\n\r\nnCr pr qn\u2013r\u00a0<\/span>\r\n\r\n \r\n\r\nWhich can be denoted B(r: n, p)\u00a0<\/span>\r\n\r\n \r\n\r\nThus P(r successes) = B(r: n, p) = nCr pr qn\u2013r r= 1, 2, 3, \u2026.., n\u00a0<\/span>\r\n\r\n \r\n\r\nIf the number X of successes is a random variable, then\u00a0<\/span>\r\n\r\n \r\n\r\nP(X = r) = B(r; n, p) = nCr pr qn\u2013r\u00a0<\/span>\r\n\r\n \r\n\r\nWe notice that the probabilities of 0 1 2 3 r n<\/span>\r\n\r\n \r\n\r\nqn nC1 p qn\u20131 nC2 p2 qn\u20132 nC3 p3 qn\u20133 nCr pr qn\u2013r pn<\/span>\r\n\r\n \r\n\r\nwhich are successive terms of the binomial expansion of <\/span>(q + p)<\/em>n<\/em>,<\/em> and hence, the above probability distribution is known as \u2018binomial distribution\u2019.<\/span>\r\n\r\n \r\n\r\nIt can be noted that the sum of all the probabilities is 1.<\/span>\r\n\r\n \r\n\r\nq<\/em>n<\/em> + <\/em>n<\/em>C<\/em>1<\/em> p q<\/em>n\u20131<\/em> + <\/em>n<\/em>C<\/em>2<\/em> p<\/em>2<\/em> q<\/em>n\u20132<\/em> + \u2026\u2026+ <\/em>n<\/em>C<\/em>r<\/em>\u00a0 p<\/em>r<\/em> q<\/em>n\u2013r<\/em> + \u2026\u2026 + p<\/em>n<\/em>\r\n\r\n \r\n\r\n=\u00a0 (q + p)n<\/span>\r\n\r\n \r\n\r\n=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1<\/span>\r\n\r\n<\/div>\r\n
      \r\n\r\n \r\n

      Example 3: <\/strong>The probability that a particular type of component will survive a given shock test is 23 .<\/em> Find the probability that<\/strong> 2 of the next 5 components tested will survive.<\/p>\r\n\r\n<\/div>\r\n

      \r\n\r\n \r\n

      Solution : <\/strong>Let shock test of the component be regarded as a trial and its survival as the success. If p<\/em> denotes the probability of success and q<\/em> that of failure in a single trial, then<\/p>\r\nNumber of tests n = 5, p =<\/em>23 , q =<\/em> 13\r\n\r\n \r\n\r\nRequired probability = p (2 successes)<\/em>\r\n\r\n= B<\/em>(2: 5, 23) = 5<\/em>C<\/em>2<\/em> (23)2 (13)3\r\n\r\n=10*4\/3*3*33*3\r\n\r\n=40\/243\r\n\r\n \r\n

      Example 4: <\/strong>The probability that a patient will survive from a delicate heart operation is 60. What is the probability that exactly 5 of the next 7 patients having this operation will survive?<\/p>\r\n \r\n\r\nSolution : <\/strong>Let the heart operation of a patient be regarded as a trial and his survival as the success.\r\n\r\nIf n<\/em> denotes the number of trials and p<\/em> probability of success and q<\/em> that of failure in a single trial then\r\n\r\n \r\n\r\nn = 7; p = 0.9, q = 1 \u2013 p = 0.1<\/em>\r\n\r\n \r\n\r\n\r\n\r\n\r\n
      Required probability = p (5 successes)<\/td>\r\n= b (5:7, 0.9)<\/td>\r\n<\/tr>\r\n
      <\/td>\r\n= 7C5 (0.9)5 (0.1)2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n7\u00d763\r\n\r\n=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0 \u00d7 0.5905 \u00d7 0.01\r\n\r\n \r\n\r\n \r\n

      Example 5: <\/strong>If 10 per cent of components produced by a machine are defective, determine the probability that out of 5 components selected at random (1) none; (2) one; (3) at most one component will be defective.<\/p>\r\n \r\n\r\nSolution : <\/strong>Let selection of a component be regarded as a trial and getting a defective component as the success.\r\n\r\n \r\n\r\nIf n<\/em> denotes the number of trials and p<\/em> the probability of success and q<\/em> that of failure,\r\n\r\n \r\n\r\nthen\r\n\r\n \r\n\r\nn = 5, p = <\/em>10010= 0.1, q = 1 \u2013 p = 0.9 <\/em>1 .P(0 success)=B (0;5,0.1)=<\/strong>(0.9)5\r\n

      2.\u00a0\u00a0 P(1 success)=B (1;5,0.1 )=<\/strong>\u00a0\u00a0\u00a0\u00a0 5C1(0.1) (0.9)4<\/p>\r\n \r\n\r\n \r\n\r\n4.\u00a0\u00a0\u00a0\u00a0\u00a0 Probability of at most 1 success = P(0 success) + P(1 success)\r\n\r\n \r\n\r\n=\u00a0 0.5905+0.3281\r\n\r\n \r\n\r\n= 0.9186\r\n\r\n \r\n\r\nExample 6: <\/strong>The incidence of an occupational disease in an industry is that the workers have 20% percent chance of suffering from it.\r\n\r\n \r\n\r\nFind the probability that out of 6 workers selected at random.\r\n\r\n(1)\u00a0\u00a0 4 will catch the diseases.<\/span>\r\n\r\n(2)\u00a0\u00a0 At least 4 will catch the disease.<\/span>\r\n\r\n<\/div>\r\n

      \r\n\r\n \r\n

      Solution: <\/strong>Let the selection of a worker be regarded as a trial and suffering from the disease as success.<\/p>\r\n \r\n\r\nIf n<\/em> denotes the number of trials and p<\/em> probability of success and q<\/em> that of failure, then n = 6, p = 10020 = 15 ,\r\n

      q = 1 \u2013 p = 1 \u2013 15= 45<\/p>\r\n \r\n\r\n(1)\u00a0<\/em>P <\/em>(4 successes) = B (4: 6, 15) = 6C4 p4 q2\r\n\r\n=\u00a06<\/em>C<\/em>4<\/em> (15)4 (45)2\r\n\r\n=\u00a06<\/em>C<\/em>4<\/em> (15)4 (45)2\r\n\r\n=\u00a0 15 x (0.2)4 (0.8)2\r\n\r\n=\u00a0 15 x 0.0016 x 0.64\r\n\r\n=\u00a0 0.01536\r\n\r\n \r\n\r\n(2)\u00a0\u00a0 <\/em>P <\/em>(at least 4 successes) = p<\/em>(4 successes) + p<\/em>(5 successes) + p<\/em>(6 successes)\r\n\r\n=\u00a0\u00a0\u00a0\u00a0 6<\/em>C<\/em>4<\/em> (15)4 (45)2 + 6<\/em>C<\/em>4<\/em>\u00a0 <\/em>(15)5 (45) + (15)6\r\n\r\n\u00a0<\/em>\r\n\r\n=\u00a0\u00a0 [ 6<\/em>C<\/em>4<\/em> (0.8)2 + 6C1 (0.2) (0.8) + (0.2)2 ]\r\n\r\n\u00a0<\/em>\r\n\r\n=\u00a0 0.0016 [15 x 0.64 + 6 x 0.16 + 0.04]\r\n\r\n \r\n\r\n=\u00a0 0.0016 [9.60 + 0.96 + 0.04]\r\n\r\n \r\n\r\n=\u00a0 0.01696\r\n\r\n \r\n\r\nExample 7:\u00a0<\/strong>The probability that a bomb dropped from a plane will strike the target is 15. If 6 bombs are dropped, find the probability that (1) exactly two will strike the target; (2) at least two will strike the target.\r\n\r\n \r\n\r\nSolution : <\/strong>Let dropping of a bomb be regarded as a trial and striking the target as the success.\r\n\r\nThen n = 6, p =<\/em> 15 , q = 1<\/em> \u2013<\/em> p =<\/em> 45\r\n\r\n(1)\u00a0p <\/em>(2 successes) = B(2: 6, 15) = 6C2 (15)2 (45)4= 15 x (0.2)2 x (0.8)4 = 0.2458\r\n\r\n(2)\u00a0p <\/em>(at least 2 success) = 1 \u2013 p<\/em>(0 success) \u2013 p <\/em>(1 success) = 1 \u2013 (45)6 \u2212 6C1 \u00d7 15 \u00d7 (45)5\r\n\r\n=\u00a0\u00a0\u00a0\u00a0 1 \u2013(45)5 (45 + 65)= 1 \u2013 2 \u00d7 (45)5\r\n\r\n=\u00a0 1 \u2013 2 x (0.8)5\r\n\r\n=\u00a0 1 \u2013 2 x 3277\r\n\r\n=1- 0.6554\r\n\r\n=0.3446\r\n\r\n<\/div>\r\n

      \r\n\r\n \r\n\r\nCOMPUTING OF CENTRAL TENDENCY AND DISPERSION FOR THE BIONOMIAL DISTRIBUTION:<\/strong>\r\n\r\n \r\n\r\nMean (\u00a0 ) = np\r\n\r\nStandard Denation (\u00a0 ) = \u221a\r\n\r\nVariance (\u00a0\u00a0 2) = npq.\r\n\r\n \r\n\r\nQ.1 <\/strong>Modern filling machines fill toothpaste to the desired level of 80% of the time. What is the probability that exactly half the tubes in a cartoon selected at random will be filled? Total tubes are six.\r\n\r\n \r\n\r\nSolution :\u00a0<\/strong>p = 0.8, q = 1 \u2013 0.8 = 0.2,\u00a0\u00a0\u00a0 n = 6.\r\n\r\n \r\n\r\nUsing Bionomical formula\r\n\r\nProbability of r success in n trials.\r\n\r\n=\u00a0 \u00a0 !px qx\u2013r\u00a0 !(\u00a0 \u2212\u00a0 )\r\n\r\n=\u00a0\u00a0\u00a0\u00a0 3!(3!)6!(0.8)3 (0.2)3\r\n\r\n=\u00a0\u00a0\u00a0\u00a0 6\u00d76720(0.512) (0.008)\r\n\r\n=\u00a0 20 (0.512) (0.008)\r\n\r\n=\u00a0 0.08192\r\n\r\n \r\n

      Q.2 <\/strong>A packaging machine produces 20 percent defective (1) packages. If we take a random sample of 10 packages. Find the mean and standard Deviation?<\/p>\r\n \r\n\r\nSolution : <\/strong>.\u00a0 \u00a0 X ~ B(n,p).\u00a0\u00a0\u00a0 n= 10, p = 0.2\r\n\r\nMean = np = 10 (0.2) = 2 \u00a0mean\r\n\r\n=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a\u00a0\u00a0 = \u221a10(0.2)(0.8) = \u221a1.6\r\n\r\n=\u00a0 1.265 \u00a0s.d.\r\n\r\n \r\n

      Q.3 <\/strong>If 5% of the items produced turn out to be defective, then find out the probability that out of 20 items selected at random then are \u2013<\/p>\r\n \r\n\r\n(a)\u00a0\u00a0\u00a0 exactly three defectives\r\n\r\n(b)\u00a0\u00a0 at least two defectives\r\n\r\n(c)\u00a0\u00a0\u00a0 exactly four defectives\r\n\r\n(d)\u00a0\u00a0 find the mean and variance.\r\n\r\n \r\n

      Solution : <\/strong>An item produced can either be defective on non defective. Thus, there are only two outcomes. The probability (p<\/em>) of a defective item being produced is 0.05 and it\u00a0remains constant. Further the trails are independent. Thus, inspecting a small sample of 20 can be conceived as a repetition of Bernauli trials. Therefore, the parameters of bionomil distribution n = 20 ; p = .05 i.e. B(n,p). Hence<\/span><\/p>\r\n\r\n<\/div>\r\n

      \r\n\r\n \r\n\r\n(a) P(x=3) = P (exactly three defective items) = (.05)3 (0.95)17\r\n\r\n= 20\u00d719\u00d718\u00d717(0.05)3 (0.95)17\r\n\r\n3\u00d72\u00d717\r\n\r\n \r\n\r\n(b)\u00a0 P(x \u2265 2) = P (at last two defective items) = Probability of two defectives + probability of three defectives + \u2026\u2026\u2026\u2026\u2026. + Probability of all defectives.\r\n\r\n \r\n\r\n=\u00a0 1 \u2013 p <\/em>(no defective) \u2013 p<\/em>(one defective).\r\n\r\n=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 - (0.95)20 - 1!9!20! (0.05)1 (0.95)1 9\r\n\r\n=\u00a0\u00a0\u00a0 0.264\r\n\r\n \r\n\r\n(c)\u00a0<\/strong>P(X = <\/strong>4)\u00a0\u00a0 = P (exactly four defectives)= 4!6!20! (0.05)4(0.95)16\r\n\r\n\u00a0<\/strong>=\u00a0 0.13\r\n\r\n \r\n\r\n(d)\u00a0 Mean\u00a0 = n p= 20 x 0.05 = 1\r\n\r\nVariance = npq\r\n\r\n= 20 x 0.05 x 0.95 = 0.95\r\n\r\n \r\n\r\nProbability of success and the shape of the Binomical distribution\r\n\r\n \r\n\r\nFor any number of trials n:\r\n\r\n \r\n\r\n1.\u00a0 The binomial probability distribution is symmetric if p = .5 0\r\n\r\n \r\n\r\n2.\u00a0 The binomial probability distribution is skewed to the right if p is less than .50\r\n\r\n \r\n\r\n3.\u00a0 \u00a0The binomial probability distribution is skewed to the left if p is greater than .50\r\n\r\n \r\n\r\nSummary<\/strong>\r\n\r\n \r\n

      Here Out of two special probability distributions (Binomial & Poisson) for a discrete random variable \u2013the binomial distribution is developed. Since decisions are made under uncertain conditions therefore, here we expain if outcomes and their probabilities for a statistical experiment are known we can find out what will happen on average if that experiment is performed many times.For a lottery example we can find out what a lottery player can expect to win (or lose) on average if he continues playing this lottery again and again.Concept of factorial & combination are also discussed.<\/p>\r\n\r\n<\/div>\r\n

      \r\n\r\n \r\n\r\nTable \u2013 1 Table of Binomial Probabilities<\/strong>\r\n\r\n<\/div>\r\n\"\"\"\"\r\n\r\n \r\n\r\nTable II- Values of<\/strong>\u00a0\u00a0\u00a0\u00a0 \u2201\u00a0 \u00a0(Combintion)<\/strong>\r\n\r\n \r\n\r\n\"\"","rendered":"
      \n

       <\/p>\n

       <\/p>\n

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

       <\/p>\n

      2.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Binomial Probability Distribution<\/p>\n

       <\/p>\n

      3.\u00a0\u00a0\u00a0\u00a0\u00a0 Binomial Experiment<\/p>\n

       <\/p>\n

      4.\u00a0\u00a0\u00a0\u00a0\u00a0 Conditions of Binomial Experiment<\/p>\n

       <\/p>\n

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

       <\/p>\n

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

       <\/p>\n

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

       <\/p>\n

      THE BINOMIAL PROBABILITY DISTRIBUTION<\/strong><\/p>\n

       <\/p>\n

      The binomial probability distribution is one of the most widely used discrete probability distributions. It is applied to find the probability that an outcome will occur x times in n performances of an experiment. For example, given that the probability is 0.5 that a mobile manufactured at a firm is defective, we may be interested in finding the probability that in a random sample of three mobiles manufactured at this firm, exactly one will be defective. As a second example, we may be interested in finding the probability that a baseball player, with a batting average of 0.250, will have no hits in 10 trips to the plate.<\/p>\n

       <\/p>\n

      To apply the binomial probability distribution, the random variable x must be a discrete dichotomous random variable. In other words, the variable must be a discrete random variable and each repetition of the experiment must result in one of two possible\u00a0outcomes. The binomial distribution is applied to experiments that satisfy the four conditions of a binomial experiment. (These conditions are described here.) Each repetition of a binomial experiment is called a trial or a Bernoulli trial (after Jacob Bernoulli). For, examples, if an experiment is defined as one toss of a coin and this experiment is repeated 10 times, then each repetition ( toss) is called a trial. Consequently, there are 10 total trials for this experiment.<\/span><\/p>\n<\/div>\n

      \n

       <\/p>\n

      THE BINOMIAL EXPERIMENT<\/strong><\/p>\n

       <\/p>\n

      An experiment that satisfies the following four conditions is called a binomial experiment.<\/p>\n

       <\/p>\n

      1.There are n identical trials, In other words, the given experiment is repeated n times. All these repetitions<\/p>\n

      are performed under identical conditions.<\/p>\n

      2. Each trial has two and only two outcomes. These outcomes are usually called a success and failure.<\/p>\n

      3. The probability of success is denoted by p and that of failure by q, and p + q = 1, The probability p and q<\/p>\n

      remain constant for each trial.<\/p>\n

      4. The trials are independent. In other words, the outcome of one trial does not affect the outcome of<\/p>\n

      another trial.<\/p>\n

       <\/p>\n

      CONDITION OF A BINOMIAL EXPERIMENT<\/strong><\/p>\n

       <\/p>\n

      A BINOMIAL EXPPERIMENT MUST SATISFY THE FOOLLOWING FOUR CONDITIONS<\/em><\/strong>.<\/p>\n

       <\/p>\n

      1.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 There are n identical trials.<\/p>\n

       <\/p>\n

      2.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Each trial has only two possible outcomes.<\/p>\n

       <\/p>\n

      3.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The probabilities of the two outcomes remain constant.<\/p>\n

       <\/p>\n

      4.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The trials are independent.<\/p>\n

       <\/p>\n

      Note that one of the two outcomes of a trial is called a success and the other a failure. Notice that a success does not mean that the corresponding outcome is considered favorable or desirable. Similarly, a failure does not necessarily refer to an unfavorable or undesirable outcome. Success and failure are simply the names used to denote the two possible outcomes of a trial. The outcome to which the question refers is usually called a success; the outcome to which it does not refer is called a failure.<\/p>\n

       <\/p>\n

      EXAMPLE 1: Consider the experiment consisting of 10 tosses of a coin. Determine if it is a binomial experiment.<\/p>\n

       <\/p>\n

      Solution; As described below, the experiment consisting of 10 tosses of a coin satisfies all four conditions of a binomial experiment.<\/p>\n

       <\/p>\n

        \n
      1. There are a total of 10 trials (tosses), and they are all identical. All 10 tosses are performed under identical conditions.<\/li>\n
      2. Each trial (toss) has only two possible outcomes: a head and a tail. Let a head be called a success and a tail be called a failure.<\/li>\n
      3. The probability of obtaining a head (a success) is \u00bd and that of a tail ( is failure) is \u00bd for any toss. That is,<\/li>\n<\/ol>\n

         <\/p>\n

        P = P(H) = \u00bd and q = P(T) =\u00a0 \u00bd<\/p>\n

         <\/p>\n

        The sum of these two probabilities is 1.0. Also, these probabilities remain the same for each toss.<\/span><\/p>\n

         <\/p>\n

        The trials (tosses) are independent. The result of any preceding toss has no bearing on the result of any succeeding toss.<\/span><\/p>\n

         <\/p>\n

        Consequently, the experiment consisting of 10 tosses is a binomial experiment.<\/span><\/p>\n<\/div>\n

        \n

         <\/p>\n

        EXAMPLE 2: Five percent of all mobiles manufactured by a large electronics company are defective. Three mobiles are randomly selected from the production line of this company. The selected mobiles are inspected to determine if each of them is defective or good. Is this experiment a binomial experiment?<\/p>\n

         <\/p>\n

        Solution<\/strong><\/p>\n

         <\/p>\n

          \n
        1. This example consists of three identical trials. A trial represents the selection of a mobile.<\/li>\n
        2. Each trial has two outcomes:\u00a0 A mobile is defective or is a mobile good.\u00a0 Let a defective mobile be called a success and a good mobile\u00a0\u00a0\u00a0 be called a failure.<\/li>\n
        3. Five percent of all mobiles are defective. So, the probability p that is a defective mobile is .05. As a result, the probability q that a mobile is good is .95. These two probabilities add up to 1.<\/li>\n
        4. Each trial (mobile) is independent. In other words, if one mobile is defective it does not affect the outcome of another mobile being defective or good. This is so because the size of the population is very large as compared to the sample size.<\/li>\n<\/ol>\n

          Since all four conditions of a binomial experiment are satisfied, this is an example of a binomial experiment.<\/p>\n

           <\/p>\n

          THE BINOMIAL PROBABILITY DISTRIBUTION AND BINOMIAL FORMULA<\/strong><\/p>\n

           <\/p>\n

          The random variable x that represents the number of successes in n trials for a binomial experiment is called a binomial random variable. The probability distribution of x in such experiments is called the binomial probability distribution or simply binomial distribution. Thus, the binomial probability distribution is applied to find the probability of x successes in n trials for a binomial experiment.<\/p>\n

           <\/p>\n

          To solve binomial problem, we determine the values of n, x, n \u2013 x, p and q then substitute these values in the binomial formula. To find the value of (n<\/em>C<\/em>r<\/em>). We can use either the combinations formula or the table of combinations (Table II)<\/p>\n

           <\/p>\n

          To find the probability of x successes in n trials for a binomial experiment, the only values needed are those of n and p. These are called the parameters of the binomial probability distribution or simply the binomial parameters. The value of q is obtained by subtracting the value of p from, 1.0 Thus, q = 1-p.<\/p>\n

           <\/p>\n

          BINOMIAL DISTRIBUTION<\/strong><\/p>\n

           <\/p>\n

          Repeated Independent Trials with two Outcomes<\/strong><\/p>\n

           <\/p>\n

          Consider an experiment of repeated trials in which each trial has only two<\/em> possible outcome. Such a situation always arises when we deal with testing of items. Each trial indicates\u00a0whether the item is defective or non-defective. We can define the two outcomes as a \u201csuccess\u201d or \u201cfailure\u201d.<\/span><\/p>\n<\/div>\n

          \n

           <\/p>\n

          These trials are independent as the outcome of any trial does not depend upon the outcome of the previous trial just as in tossing of a coin or drawing a card in succession from a pack of card if the card is replaced every time before the next draw. These independent repeated trials are called Bernoulli trials named after swiss mathematician Jacob Bernoulli. Further the probability of success or that of failure remains constant in each trial.<\/p>\n

           <\/p>\n

          Let a trial be repeated so that we have a series if n<\/em> trials. Let p<\/em> be the Probability of success and q<\/em> that of failure in each Bernoulli trial so that p + q = 1.<\/em><\/p>\n

           <\/p>\n

          It is required to find the probability of r<\/em> successes in n<\/em> trials. If S<\/em> denotes the success and F<\/em> the failure, then the probability of r<\/em> successes and (n<\/em> \u2013<\/em> r<\/em>) failures in a given order say.<\/p>\n

           <\/p>\n

          SFSSFS \u2026..<\/span><\/p>\n

          is pqppqp \u2026\u2026 = pr qn\u2013r<\/p>\n

          as the number is p\u2019s<\/em> is equal to the number of successes, that is, r<\/em> and the number of q\u2019s <\/em>is equal to the number of failures, that is, n \u2013 r<\/em>.<\/p>\n

           <\/p>\n

          In order to find the required probability, we have to count all different orders, in which r<\/em> successes and (n<\/em> \u2013<\/em><\/p>\n

          r<\/em>) failures can occur. This is in fact all possible arrangements of n<\/em> letters.<\/p>\n<\/div>\n

          \n

          Which r are S and (n \u2013 r) are F and is equal to !\/ !( \u2212 )! i.e.<\/p>\n

          Hence, by addition theorem, the probability of getting r<\/em> successes in a trial is<\/p>\n

           <\/p>\n

          nCr pr qn\u2013r\u00a0<\/span><\/p>\n

           <\/p>\n

          Which can be denoted B(r: n, p)\u00a0<\/span><\/p>\n

           <\/p>\n

          Thus P(r successes) = B(r: n, p) = nCr pr qn\u2013r r= 1, 2, 3, \u2026.., n\u00a0<\/span><\/p>\n

           <\/p>\n

          If the number X of successes is a random variable, then\u00a0<\/span><\/p>\n

           <\/p>\n

          P(X = r) = B(r; n, p) = nCr pr qn\u2013r\u00a0<\/span><\/p>\n

           <\/p>\n

          We notice that the probabilities of 0 1 2 3 r n<\/span><\/p>\n

           <\/p>\n

          qn nC1 p qn\u20131 nC2 p2 qn\u20132 nC3 p3 qn\u20133 nCr pr qn\u2013r pn<\/span><\/p>\n

           <\/p>\n

          which are successive terms of the binomial expansion of <\/span>(q + p)<\/em>n<\/em>,<\/em> and hence, the above probability distribution is known as \u2018binomial distribution\u2019.<\/span><\/p>\n

           <\/p>\n

          It can be noted that the sum of all the probabilities is 1.<\/span><\/p>\n

           <\/p>\n

          q<\/em>n<\/em> + <\/em>n<\/em>C<\/em>1<\/em> p q<\/em>n\u20131<\/em> + <\/em>n<\/em>C<\/em>2<\/em> p<\/em>2<\/em> q<\/em>n\u20132<\/em> + \u2026\u2026+ <\/em>n<\/em>C<\/em>r<\/em>\u00a0 p<\/em>r<\/em> q<\/em>n\u2013r<\/em> + \u2026\u2026 + p<\/em>n<\/em><\/p>\n

           <\/p>\n

          =\u00a0 (q + p)n<\/span><\/p>\n

           <\/p>\n

          =\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1<\/span><\/p>\n<\/div>\n

          \n

           <\/p>\n

          Example 3: <\/strong>The probability that a particular type of component will survive a given shock test is 23 .<\/em> Find the probability that<\/strong> 2 of the next 5 components tested will survive.<\/p>\n<\/div>\n

          \n

           <\/p>\n

          Solution : <\/strong>Let shock test of the component be regarded as a trial and its survival as the success. If p<\/em> denotes the probability of success and q<\/em> that of failure in a single trial, then<\/p>\n

          Number of tests n = 5, p =<\/em>23 , q =<\/em> 13<\/p>\n

           <\/p>\n

          Required probability = p (2 successes)<\/em><\/p>\n

          = B<\/em>(2: 5, 23) = 5<\/em>C<\/em>2<\/em> (23)2 (13)3<\/p>\n

          =10*4\/3*3*33*3<\/p>\n

          =40\/243<\/p>\n

           <\/p>\n

          Example 4: <\/strong>The probability that a patient will survive from a delicate heart operation is 60. What is the probability that exactly 5 of the next 7 patients having this operation will survive?<\/p>\n

           <\/p>\n

          Solution : <\/strong>Let the heart operation of a patient be regarded as a trial and his survival as the success.<\/p>\n

          If n<\/em> denotes the number of trials and p<\/em> probability of success and q<\/em> that of failure in a single trial then<\/p>\n

           <\/p>\n

          n = 7; p = 0.9, q = 1 \u2013 p = 0.1<\/em><\/p>\n

           <\/p>\n\n\n\n\n
          Required probability = p (5 successes)<\/td>\n= b (5:7, 0.9)<\/td>\n<\/tr>\n
          <\/td>\n= 7C5 (0.9)5 (0.1)2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

           <\/p>\n

          7\u00d763<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0 \u00d7 0.5905 \u00d7 0.01<\/p>\n

           <\/p>\n

           <\/p>\n

          Example 5: <\/strong>If 10 per cent of components produced by a machine are defective, determine the probability that out of 5 components selected at random (1) none; (2) one; (3) at most one component will be defective.<\/p>\n

           <\/p>\n

          Solution : <\/strong>Let selection of a component be regarded as a trial and getting a defective component as the success.<\/p>\n

           <\/p>\n

          If n<\/em> denotes the number of trials and p<\/em> the probability of success and q<\/em> that of failure,<\/p>\n

           <\/p>\n

          then<\/p>\n

           <\/p>\n

          n = 5, p = <\/em>10010= 0.1, q = 1 \u2013 p = 0.9 <\/em>1 .P(0 success)=B (0;5,0.1)=<\/strong>(0.9)5<\/p>\n

          2.\u00a0\u00a0 P(1 success)=B (1;5,0.1 )=<\/strong>\u00a0\u00a0\u00a0\u00a0 5C1(0.1) (0.9)4<\/p>\n

           <\/p>\n

           <\/p>\n

          4.\u00a0\u00a0\u00a0\u00a0\u00a0 Probability of at most 1 success = P(0 success) + P(1 success)<\/p>\n

           <\/p>\n

          =\u00a0 0.5905+0.3281<\/p>\n

           <\/p>\n

          = 0.9186<\/p>\n

           <\/p>\n

          Example 6: <\/strong>The incidence of an occupational disease in an industry is that the workers have 20% percent chance of suffering from it.<\/p>\n

           <\/p>\n

          Find the probability that out of 6 workers selected at random.<\/p>\n

          (1)\u00a0\u00a0 4 will catch the diseases.<\/span><\/p>\n

          (2)\u00a0\u00a0 At least 4 will catch the disease.<\/span><\/p>\n<\/div>\n

          \n

           <\/p>\n

          Solution: <\/strong>Let the selection of a worker be regarded as a trial and suffering from the disease as success.<\/p>\n

           <\/p>\n

          If n<\/em> denotes the number of trials and p<\/em> probability of success and q<\/em> that of failure, then n = 6, p = 10020 = 15 ,<\/p>\n

          q = 1 \u2013 p = 1 \u2013 15= 45<\/p>\n

           <\/p>\n

          (1)\u00a0<\/em>P <\/em>(4 successes) = B (4: 6, 15) = 6C4 p4 q2<\/p>\n

          =\u00a06<\/em>C<\/em>4<\/em> (15)4 (45)2<\/p>\n

          =\u00a06<\/em>C<\/em>4<\/em> (15)4 (45)2<\/p>\n

          =\u00a0 15 x (0.2)4 (0.8)2<\/p>\n

          =\u00a0 15 x 0.0016 x 0.64<\/p>\n

          =\u00a0 0.01536<\/p>\n

           <\/p>\n

          (2)\u00a0\u00a0 <\/em>P <\/em>(at least 4 successes) = p<\/em>(4 successes) + p<\/em>(5 successes) + p<\/em>(6 successes)<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0 6<\/em>C<\/em>4<\/em> (15)4 (45)2 + 6<\/em>C<\/em>4<\/em>\u00a0 <\/em>(15)5 (45) + (15)6<\/p>\n

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

          =\u00a0\u00a0 [ 6<\/em>C<\/em>4<\/em> (0.8)2 + 6C1 (0.2) (0.8) + (0.2)2 ]<\/p>\n

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

          =\u00a0 0.0016 [15 x 0.64 + 6 x 0.16 + 0.04]<\/p>\n

           <\/p>\n

          =\u00a0 0.0016 [9.60 + 0.96 + 0.04]<\/p>\n

           <\/p>\n

          =\u00a0 0.01696<\/p>\n

           <\/p>\n

          Example 7:\u00a0<\/strong>The probability that a bomb dropped from a plane will strike the target is 15. If 6 bombs are dropped, find the probability that (1) exactly two will strike the target; (2) at least two will strike the target.<\/p>\n

           <\/p>\n

          Solution : <\/strong>Let dropping of a bomb be regarded as a trial and striking the target as the success.<\/p>\n

          Then n = 6, p =<\/em> 15 , q = 1<\/em> \u2013<\/em> p =<\/em> 45<\/p>\n

          (1)\u00a0p <\/em>(2 successes) = B(2: 6, 15) = 6C2 (15)2 (45)4= 15 x (0.2)2 x (0.8)4 = 0.2458<\/p>\n

          (2)\u00a0p <\/em>(at least 2 success) = 1 \u2013 p<\/em>(0 success) \u2013 p <\/em>(1 success) = 1 \u2013 (45)6 \u2212 6C1 \u00d7 15 \u00d7 (45)5<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0 1 \u2013(45)5 (45 + 65)= 1 \u2013 2 \u00d7 (45)5<\/p>\n

          =\u00a0 1 \u2013 2 x (0.8)5<\/p>\n

          =\u00a0 1 \u2013 2 x 3277<\/p>\n

          =1- 0.6554<\/p>\n

          =0.3446<\/p>\n<\/div>\n

          \n

           <\/p>\n

          COMPUTING OF CENTRAL TENDENCY AND DISPERSION FOR THE BIONOMIAL DISTRIBUTION:<\/strong><\/p>\n

           <\/p>\n

          Mean (\u00a0 ) = np<\/p>\n

          Standard Denation (\u00a0 ) = \u221a<\/p>\n

          Variance (\u00a0\u00a0 2) = npq.<\/p>\n

           <\/p>\n

          Q.1 <\/strong>Modern filling machines fill toothpaste to the desired level of 80% of the time. What is the probability that exactly half the tubes in a cartoon selected at random will be filled? Total tubes are six.<\/p>\n

           <\/p>\n

          Solution :\u00a0<\/strong>p = 0.8, q = 1 \u2013 0.8 = 0.2,\u00a0\u00a0\u00a0 n = 6.<\/p>\n

           <\/p>\n

          Using Bionomical formula<\/p>\n

          Probability of r success in n trials.<\/p>\n

          =\u00a0 \u00a0 !px qx\u2013r\u00a0 !(\u00a0 \u2212\u00a0 )<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0 3!(3!)6!(0.8)3 (0.2)3<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0 6\u00d76720(0.512) (0.008)<\/p>\n

          =\u00a0 20 (0.512) (0.008)<\/p>\n

          =\u00a0 0.08192<\/p>\n

           <\/p>\n

          Q.2 <\/strong>A packaging machine produces 20 percent defective (1) packages. If we take a random sample of 10 packages. Find the mean and standard Deviation?<\/p>\n

           <\/p>\n

          Solution : <\/strong>.\u00a0 \u00a0 X ~ B(n,p).\u00a0\u00a0\u00a0 n= 10, p = 0.2<\/p>\n

          Mean = np = 10 (0.2) = 2 \u00a0mean<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a\u00a0\u00a0 = \u221a10(0.2)(0.8) = \u221a1.6<\/p>\n

          =\u00a0 1.265 \u00a0s.d.<\/p>\n

           <\/p>\n

          Q.3 <\/strong>If 5% of the items produced turn out to be defective, then find out the probability that out of 20 items selected at random then are \u2013<\/p>\n

           <\/p>\n

          (a)\u00a0\u00a0\u00a0 exactly three defectives<\/p>\n

          (b)\u00a0\u00a0 at least two defectives<\/p>\n

          (c)\u00a0\u00a0\u00a0 exactly four defectives<\/p>\n

          (d)\u00a0\u00a0 find the mean and variance.<\/p>\n

           <\/p>\n

          Solution : <\/strong>An item produced can either be defective on non defective. Thus, there are only two outcomes. The probability (p<\/em>) of a defective item being produced is 0.05 and it\u00a0remains constant. Further the trails are independent. Thus, inspecting a small sample of 20 can be conceived as a repetition of Bernauli trials. Therefore, the parameters of bionomil distribution n = 20 ; p = .05 i.e. B(n,p). Hence<\/span><\/p>\n<\/div>\n

          \n

           <\/p>\n

          (a) P(x=3) = P (exactly three defective items) = (.05)3 (0.95)17<\/p>\n

          = 20\u00d719\u00d718\u00d717(0.05)3 (0.95)17<\/p>\n

          3\u00d72\u00d717<\/p>\n

           <\/p>\n

          (b)\u00a0 P(x \u2265 2) = P (at last two defective items) = Probability of two defectives + probability of three defectives + \u2026\u2026\u2026\u2026\u2026. + Probability of all defectives.<\/p>\n

           <\/p>\n

          =\u00a0 1 \u2013 p <\/em>(no defective) \u2013 p<\/em>(one defective).<\/p>\n

          =\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 – (0.95)20 – 1!9!20! (0.05)1 (0.95)1 9<\/p>\n

          =\u00a0\u00a0\u00a0 0.264<\/p>\n

           <\/p>\n

          (c)\u00a0<\/strong>P(X = <\/strong>4)\u00a0\u00a0 = P (exactly four defectives)= 4!6!20! (0.05)4(0.95)16<\/p>\n

          \u00a0<\/strong>=\u00a0 0.13<\/p>\n

           <\/p>\n

          (d)\u00a0 Mean\u00a0 = n p= 20 x 0.05 = 1<\/p>\n

          Variance = npq<\/p>\n

          = 20 x 0.05 x 0.95 = 0.95<\/p>\n

           <\/p>\n

          Probability of success and the shape of the Binomical distribution<\/p>\n

           <\/p>\n

          For any number of trials n:<\/p>\n

           <\/p>\n

          1.\u00a0 The binomial probability distribution is symmetric if p = .5 0<\/p>\n

           <\/p>\n

          2.\u00a0 The binomial probability distribution is skewed to the right if p is less than .50<\/p>\n

           <\/p>\n

          3.\u00a0 \u00a0The binomial probability distribution is skewed to the left if p is greater than .50<\/p>\n

           <\/p>\n

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

           <\/p>\n

          Here Out of two special probability distributions (Binomial & Poisson) for a discrete random variable \u2013the binomial distribution is developed. Since decisions are made under uncertain conditions therefore, here we expain if outcomes and their probabilities for a statistical experiment are known we can find out what will happen on average if that experiment is performed many times.For a lottery example we can find out what a lottery player can expect to win (or lose) on average if he continues playing this lottery again and again.Concept of factorial & combination are also discussed.<\/p>\n<\/div>\n

          \n

           <\/p>\n

          Table \u2013 1 Table of Binomial Probabilities<\/strong><\/p>\n<\/div>\n

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

           <\/p>\n

          Table II- Values of<\/strong>\u00a0\u00a0\u00a0\u00a0 \u2201\u00a0 \u00a0(Combintion)<\/strong><\/p>\n

           <\/p>\n

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