"There are 10 contestants in an archery competition. 4 of the contestants are women. Prizes are awarded to the top 4 competitors. What is the probability that at least one woman finishes in the first 4 places."
I've been going around in circles a little bit with this one, as it is from a draft textbook that has quite a few incorrect answers. I get:
$P_4^{10}=5040$ for the total ways of arranging the top 4.
$P_4^6=360$ for the number of ways of arranging a top 4 with no women.
$P_4^{10}-P_4^6=4680$ for the number of ways of arranging a top 4 with at least one woman.
So the probability of at least one woman finishing in the top 10 is $4680/5040$ $=13/14$
However, the text book says $1/14$.
Apologies if I'm missing something very obvious.
You are definately right. The probability that there's no woman in the top 4 (let's call it $A$) is $$P(A) = \frac{6*5*4*3}{10*9*8*7}=\frac{1}{14}.$$ So the probability that there's at least one woman in the top 4 is $$P(A')=1-P(A)=\frac{13}{14}.$$