A salesperson has 15 major accounts Eight of those accounts

A salesperson has 15 major accounts. Eight of those accounts were sent the wrong bill last month (billed too much by the dishonest account we discussed in class). The salesperson want s to call her accounts and explain what happened, but she only has time to call 10 of them before quitting time.   Assume that a call to a wrongly billed account takes as much time as a call to a correctly billed account, and that she will not stay past quitting time in order to call all 15. If she selects the accounts and calls them randomly, find the probability that she will reach:

         a.       6 of the 8 she wants to contact.

         b.      only 5 she wants to contact.

         c.      none of those incorrectly billed

Solution

A)

Note that the probability of x successes out of n trials is          
          
P(x) = C(N-K, n-x) C(K, x) / C(N, n)          
          
where          
N = population size =    15      
K = number of successes in the population =    8      
n = sample size =    10      
x = number of successes in the sample =    6      
          
Thus,          
          
P(   6   ) =    0.326340326 [answer]

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B)

Note that the probability of x successes out of n trials is          
          
P(x) = C(N-K, n-x) C(K, x) / C(N, n)          
          
where          
N = population size =    15      
K = number of successes in the population =    8      
n = sample size =    10      
x = number of successes in the sample =    5      
          
Thus,          
          
P(   5   ) =    0.391608392 [ANSWER]

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c)

This is impossible, since there are only 7 that are correctly billed. Therefore, he is pineed to call at least 3 that are incorrectly billed.

Thus,

P(0) = 0 [ANSWER]