2009 AIME I Problem 3

Question

A coin that comes up heads with probability

p \gt 0

and tails with probability

1 - p \gt 0

independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to

\frac{1}{25}

of the probability of five heads and three tails. Let

p = \frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Set up

\binom{8}{3}p^3(1-p)^5 = \frac{1}{25}\binom{8}{5}p^5(1-p)^3

The binomial coefficients cancel, leaving

(1-p)^2 = \frac{p^2}{25};

both sides are positive, so take square roots

The condition says

\begin{aligned} &\binom{8}{3} p^3 (1-p)^5 \\ &= \frac{1}{25} \binom{8}{5} p^5 (1-p)^3. \end{aligned}

Since

\binom{8}{3} = \binom{8}{5}

and both

p

and

1 - p

are positive, dividing by

p^3(1-p)^3

leaves

(1-p)^2 = \frac{p^2}{25},

so

1 - p = \frac{p}{5}.

Hence

p = \frac{5}{6},

and

m + n = 5 + 6 = 11.

Problem 3 in other years