Question

This problem has been solved!

Answer 1

Answer:

a) The probability that all the cordless phones are among the first fourteen to be serviced is 0.0295.

b) The probability that after servicing fourteen of these phones, phones of only two of the three types remain to be serviced is 0.0885.

c) The probability that two phones of each type are among the first six serviced is 0.1707.

Step-by-step explanation:

Twenty one telephones are arranged in order in which they will be serviced. Seven of these telephones are cellular, seven are cordless, and the other seven are corded phones.

Let p denotes the probability .

Out of 21, first fourteen phones to be serviced can be selected in  21 C 14  ways.

Out of those 14, seven cordless phones can be selected in  14 C 7  ways.

a) The probability that all the cordless phones are among the first fourteen to be serviced is given by:

                           p  =  (14 C 7 )*(21 C 14 ) $=0.0295$

b)  Let the probability that after servicing fourteen of these phones, phones of only two of the three types remain to be serviced be denoted by s and is same as choosing one of three types of phones to be serviced as two phones will be remaining for the service.

Out of three type one type is selected in { }^3{C-1} ways and the probability of choosing 14 phones of which one is of same type is p.

                          q= (3 C 1)  * p  =  3  *  0.0295  =  0.0885

The probability that after servicing fourteen of these phones, phones of only two of the three types remain to be serviced is 0.0885.

c) The probability that two phones of each type are among the first six serviced is given by:

                  p  = ( 7 C 2)  * ( 7 C 2)  * (7 C 2)  * (21 C 6)

                     =  0.1707

The probability that two phones of each type are among the first six serviced is 0.1707.