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Ten boys and girls write their names on slips of paper-one name per slip-to enter a prize drawing. Two of the names are drawn at random without replacement. If the probability that both winners are boys is $\frac1{15}$, how many boys are in the group?

I started off by noting that there are $10$ people in total, and tried multiplying it with $\frac1{15}$ and I got that there were $\frac{2}{5}$th but that's not possible, please help. thank you.

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Let there be $b$ boys in the group of $10$ students. Total number of ways of choosing $2$ out of $10$ is $\binom{10}{2}$. Number of favorable cases i.e. the number of ways of choosing $2$ boys out of $b$ boys is $\binom{b}{2}$. Hence, the desired probability is $$\frac{\binom{b}{2}}{\binom{10}{2}} = \frac{b(b-1)}{10 \times 9}$$ Solve the quadratic assuming that $b$ is non-negative to get the number of boys.

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im sorry im a little confused on the notation... for the b and 10 above the 2, is the b and 10 being divided by the 2? or what does that mean? – user8051 Apr 19 '11 at 23:53
It's a binomial coefficient (Google it!), $b$-choose-$2$. – Gerry Myerson Apr 20 '11 at 1:13
@uer8051: It is the binomial coefficient. $\binom{n}{r}$ denotes the number of ways of choosing $r$ objects from $n$ objects. $\binom{n}{r} = \frac{n!}{r! (n-r)!}$ – user17762 Apr 20 '11 at 1:19
oh ok. thank you! =) – user8051 Apr 20 '11 at 23:27

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