Question

An advertising agency has 5 people (2 men and 3 women) for customer service. In waiting they have to serve 4 four clients, 2 of which are very difficult to deal with. To decide which of the employees performs the first care, select a person at random.

a) What is the probability that the first attention will be performed by a woman?

b) What is the probability that the first attention is performed by a man?

c) What is the probability that the first client served will be difficult to deal with?

d) What is the probability that the second client was attended by a woman given that the first client was attended by a man?

e) What is the probability that the second client served is difficult to treat given that the first client served is difficult to treat?

f) What is the probability that the first client is attended by a woman, the second by a man, the third by a man and the fourth by a woman?

Thanks for your help

Answer 1

a. 3/5

b.2/5(2 out of 5 are men) similar to the 1st one

c.2/4(2out of4 are difficult to deal with)

d.1 man out of 5 is gone now the next will get 4 choices( 1- man 3- women)

ans= 3/4

e. if one client is gone and also he is difficult to treat, we have 3 clients with 1 difficult to treat

ans.=1/3

f. (P)1st client attending a woman= 3/5

(P) 2nd client attending a man=2/4

(P)3 rd by a man=1/3

(P)4th by woman = 2/2=1

total (P) for everything to happen in this order=(3/5)(2/4)(1/3)(1)=1/10

ans. 1/10