Data management!! [ Hi economyst] Please Help.
posted by rock johnny .
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 nchoosex is n!/x!*(nx)! 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 (1P) where P is the probability of picking 3 females.@@@@
I would be really greatful if you help me solve it. Thanks again. :)

Out of the 14 tryouts, he picks 3. The possible combinations are 14choose3 = 14!/3!*(143)! = (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 9choose2 possible picks of the males. and 4choose1 females. Following the nchoosex 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 (1P) where P is the probability of getting all female. The number of ways to pick 3 females out of the 4 available is 4choose3 = 4!/3!(43)! = 4. So P= 4/364.