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Unit 14: Theoretical Probability Distributions



            Let the event that an item is found to be defective be termed as a success. Thus, we are given n =  Notes
            12 and p = 0.1.
                  a       12      0    12
            (i)  P r = 0f = C 0.1 a f 0.9 a f = 0.2824
                             0
            (ii) a        12  1   1    11
                 P r =1f = C 0.1 a f 0.9 a f = 0.3766
                  a       12      2    10
            (iii) P r = 2f = C 0.1 a f 0.9 a f = 0.2301
                             2
            (iv)  P (r   ) 2 = P(r = 0) + P(r = 1) +P(r = 2)
                          = 0.2824 + 0.3766 + 0.2301 = 0.8891
            (v)  P(r  2) = 1 - P(0) - P(1) = 1 - 0.2824 - 0.3766 = 0.3410


                 Example 7: In a large group of students 80% have a recommended statistics book. Three
            students are selected at random. Find the probability distribution of the number of students
            having the book. Also compute the mean and variance of the distribution.
            Solution.

            Let the event that 'a student selected at random has the book' be termed as a success. Since the
            group of students is large, 3 trials, i.e., the selection of 3 students, can be regarded as independent
            with probability of a success p = 0.8. Thus, the conditions of the given experiment satisfies the
            conditions of binomial  distribution.
            The probability mass function  P r a f = C 0.8 a f 0.2 a f 3- r
                                                  r
                                           3
                                             r            ,
            where r = 0, 1, 2 and 3
            The mean is np = 3 ´  0.8 = 2.4 and Variance is npq = 2.4 ´  0.2 = 0.48


                   Example 8:
            (a)  The mean and variance of a discrete random variable X are 6 and 2 respectively. Assuming
                 X to be a binomial variate, find P(5  X  7).

            (b)  In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2
                 successes are 0.4096 and 0.2048 respectively. Calculate the mean, variance and mode of the
                 distribution.
            Solution.
            (a)  It is given that np = 6 and npq = 2

                       npq   2   1             1   2            3
                                            1
                                                             6
                   q =    =   =   so that  p =   =   and n = ´  =  9
                       np    6   3             3   3            2
                 Now P(5  X  7) = P(X = 5) + P(X = 6) +P(X = 7)

                                 5  1  4        6  1  3         7  1  2
                                            æ
                                                             2 ö æ ö
                                                            æ
                                              2ö æ ö
                              2ö æ ö
                             æ
                        =  9 C 5 ç ÷ ç ÷  +  9 C 6 ç ÷ ç ÷  +  9 C 7 ç ÷ ç ÷
                             è
                                                            è
                                              3 ø è ø
                                                             3 ø è ø
                              3 ø è ø
                                            è
                                                   3
                                   3
                                                                  3
                         2 5  9    9        9        2 5
                        =  9 ë é  C +  C ´  2 +  C ´  4ù =  9  ´  438
                                                  û
                                     6
                                5
                                              7
                         3                           3
            (b)  Let p be the probability of a success. It is given that
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