Question

a consignment of 15 record players contain 4 defectives .the record player are selected at random one by one and examined. those examined are not put back. calculate the probability that the 9th one examined is the last defective​

Answer 1

Given:

A consignment of 15 record players contain 4 defectives .the record player are selected at random one by one and examined. those examined are not put back.  

To find:

Calculate the probability that the 9th one examined is the last defective​

Solution:

From given, we have,

The first  8  record players examined contain the remaining  3  detectives. (as 9th one examined is the last defective​)

The number of ways in which we can have  3  defectives in a total of  8  objects is  C(8,3) = 8!/3!(8 − 3)! = 56.

The number of ways in which we can have  4  defectives in a total of  15  objects is  C(15,4) = 15!/4!(15 − 4)! = 1365.

The probability that the ninth one examined is the last defective  = 56/1365 = 8/195

Therefore, the probability that the 9th one examined is the last defective​ is 8/195