Question

Hi. economyst. Thanks for your help with my other questions. But I am still stuck with this question:

@@@@My question: Rain is an amazing singer, but cannot play any instruments. He desperately wants a rock band that has a drummer, a guitarist, a bass player, and himself on vocals. He decides to post an ad at the local coffee house for open auditions for his band - The Combinations.
At the auditions, Rain gets 5 females and 9 males trying out for the remaining 3 positions in the band that can play all the instruments.

a) What is the probability that the band will have 3 males and 1 female?

b) What is the probability that the band will have at least 2 males?@@@@

@@@@economyst answered:

* Another Data Management Question - economyst, Saturday, September 26, 2009 at 11:02am

Follow the logic i gave you in your previous post. Remember, the formula for n-choose-x is n!/x!*(n-x)! where ! means factorial.

* Another Data Management Question - economyst, Saturday, September 26, 2009 at 1:43pm

hint for b) Since Rain is male, having at least two males means picking at least one male. Which is equal to (1-P) where P is the probability of picking 3 females.@@@@

I would be really greatful if you help me solve it. Thanks again. :)

Answer 1

Out of the 14 try-outs, he picks 3. The possible combinations are 14-choose-3 = 14!/3!*(14-3)! = (12*13*14)/(1*2*3) = 364

a) to get 3 M and 1 F, besides himself he needs to pick 2 M and 1 F. There are 9-choose-2 possible picks of the males. and 4-choose-1 females. Following the n-choose-x formula:
8*9/2 = 36 possible male picks and
4/1 = 4 female.
So the total possible ways to pick 2 males and 1 female is 36*4=144, The probability is therefore 144/364

b) From my hint, the prob of getting at least 2 Males is (1-P) where P is the probability of getting all female. The number of ways to pick 3 females out of the 4 available is 4-choose-3 = 4!/3!(4-3)! = 4. So P= 4/364.