A Family Decides To Have 14 Children. Find The Probability Of:
A) No Girls
B) Exactly 1 Boy
C) At Least 1 Girl
D) At Most 13 Boys

Please Help!!

P(Boy) = 0.5
P(Girl) = 0.5

A) 14 boys. (0.5)^14
B) 13 girls, 1 boy. (0.5)(0.5)^13
C) instead of doing all of these

P(at least 1 girl) = 1 - P(No girls) = 1 - P(all boys) = 1 - (0.5)^14

D) at most 13 boys. So... P(0 boy) + P(1 boy) +... + P(13 boys)
which is equal to 1 - P(14 boys) = 1 - (0.5)^14

someone check this math for me. I've been up for too long.

ISF: You're answer to B is incorrect. What you show is the probability of the first child being a boy then being followed by 13 girls in a row. But what about the case of the boy being the 2nd child, or the 3rd, etc? Since the boy can be in any one of 14 positions, the correct answer is 14*(0.5)*(0.5)^13

Good call, I knew I was missing something. Thanks.