Question

A team of 4 members is chosen from a team of 3 trainees, 3 engineers and 5 managers. find the probability that exactly 3 of them are managers?

Answer 1

Answer:

Probability =$\frac{3}{11}$

Step-by-step explanation:

Given : A team of 4 members is chosen.

From a team 3 trainees, 3 engineers and 5 managers.

To find : The probability that exactly 3 of them are managers.

Solution :

Total number of outcomes = 3+3+5 = 11

We have to choose 4 members out of 11 = $^{11}{C}_4=\frac{11!}{4!\times7!}=330$

Exactly 3 are managers out of 5 =  $^{5}{C}_3=\frac{5!}{3!\times2!}=10$    

Now, 1 member is choosen from 6(3 trainee and 3 engineers)=  $^{6}{C}_1=\frac{6!}{1!\times5!}=6$    

Probability of 4 members is $\frac{^{5}{C}_3\times^{6}{C}_1}{^{11}{C}_4}$

Probability =  $\frac{10\times6}{330}=\frac{3}{11}$