From a group of 6 boys and 6 girls a volleyball team of 6 players is

Order is not important hence combinations can be used.

(6C4)*(6C2) / (12C6)
= E

P(4B2G)=FAVORABLE CASES/POSSIBLE CASES
TOTAL: 12C6
FAVORABLE: 6C4*6C2
P(4B2G)=6C4*6C2/12C6=75/308

Ans (E)

The number of ways to select 2 girls from 6 girls is 6C2 = (6 x 5)/2! = 15. Similarly, the number of ways to select 4 boys from 6 boys is 6C4 = (6 x 5 x 4 x 3)/(4 x 3 x 2) = 15. Therefore, the total number of ways of selecting 2 girls and 4 boys is 15 x 15 = 225.

The number of ways to select 6 players from 12 players is:

12C6 = (12 x 11 x 10 x 9 x 8 x 7)/(6 x 5 x 4 x 3 x 2) = 2 x 11 x 2 x 3 x 7 = 924

So, the overall probability is 225/924 = 75/308.

Alternate Solution:

Let’s first calculate the probability of the choice B - B - B - B - G - G, in that order:

6/12 x 5/11 x 4/10 x 3/9 x 6/8 x 5/7 = (6 x 5 x 4 x 3 x 6 x 5)/(12 x 11 x 10 x 9 x 8 x 7) = 5/(11 x 4 x 7) = 5/308

Notice that B - B - B - B - G - G can be arranged in 6!/(4!*2!) = (6 x 5)/2 = 15 ways. Thus, the probability that the team consists of 4 boys and 2 girls is (5/308) x 15 = 75/308.

Answer: E

I have a stupid question. Why isn't the total outcome over here = 2^ 6 = 64 ?

Isn't this similar to our fair coin problems where we have 2 possible outcomes for one scenario and we can just multiply them together.

Thank you for the feedback !

is there a simpler way to solve this question