Question

Satish and Gaurav appear in an interview for two posts in a company. The
probability of selection of Gaurav is 0.6 and that of Satish is 0.2

(i) The probability that only Gaurav is selected:
(a) 0.48 (b) 0.56 (c) 0.32 (d) 0.08

(ii) The probability that only one of them is selected:
(a) 0.48 (b) 0.56 (c) 0.32 (d) 0.08

(iii) The probability that none of them is selected:
(a) 0.48 (b) 0.56 (c) 0.32 (d) 0.08

(iv) The probability that both of them are selected :
(a) 0.48 (b) 0.56 (c) 0.12 (d) 0.08

(v) The probability that only Satish is selected :
(a) 0.48 (b) 0.56 (c) 0.32 (d) 0.08

Answer 1

Given : Satish and Gaurav appear in an interview for two posts in a company. The probability of selection of Gaurav is 0.6 and that of Satish is 0.2

To Find :

(i) The probability that only Gaurav is selected:

(a) 0.48 (b) 0.56 (c) 0.32 (d) 0.08

(ii) The probability that only one of them is selected:

(a) 0.48 (b) 0.56 (c) 0.32 (d) 0.08

(iii) The probability that none of them is selected:

(a) 0.48 (b) 0.56 (c) 0.32 (d) 0.08

(iv) The probability that both of them are selected :

(a) 0.48 (b) 0.56 (c) 0.12 (d) 0.08

(v) The probability that only Satish is selected :

(a) 0.48 (b) 0.56 (c) 0.32 (d) 0.08

Solution :

The probability of selection of Gaurav is P(G) = 0.6

=> probability of not selecting  Gaurav is P( not G) =  0. 4   ( 1 - 0.6 = 0.4)

probability of selection of Satish is  P(S)= 0.2

probability of not selecting  Satish is P( not S) =  0. 8   ( 1 - 0.2 = 0.8)

The probability that only Gaurav is selected:

P(G) P( not S)

= ( 0.6)(0.8)

= 0.48

The probability  that only one of them is selected:

=  P(G) P( not S) + P(not G)P(S)

=   ( 0.6)(0.8)  +  ( 0.4)(0.2)

= 0.48 + 0.08

= 0.56

The probability that none of them is selected:

= P(not G) P ( not S)

= (0.4) * (0.8)

= 0.32

The probability that both of them are selected :

=P(G)P(S)

= 0.6 * 0.2

= 0.12

probability that only Satish is selected :

P(notG)P(S)

0.4 * 0.2

=0.08

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