so we start out by seeing how many ways we can select 4 out of 24, which is:
so for all male we would have:
for all females it would be:
at least 1 male is:
at least 1 female:
Hi folks!
I've this one to solve, but I just can't figure it. It is probably quite simple however.
A 4 person leadership committee is randomly chosen from a group of 24 candidates. Ten of the candidates are men, and 14 are women. Find the probability that:
A. The committe is all male or all female
B. The commitee has at least 1 man or at least 1 woman
I know this is using mutually excluisive/inclusive events, and possibly a permutation/combination thing somewhere.
Thanks in advance folks!
Right so I get everything else but for the "at least one male or at least one female" bit--- how would you solve that?
I suppose one would have to find the probability of both happening.
Would that be had by multiplying 125/138 by 248/253 ?
so P(at least one man or at least one woman) = 125/138 + 248/253 - 31000/34914 = approx. 0.9981
Right?