Question

When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac{i}{j},$ in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j.$

Answer

Answer: 283

Concepts:binomial probability fraction linear equation

Difficulty rating: 2110

Small Hint:

Let $p$ be the probability of heads and equate the two binomial probabilities

Big Hint:

Cancel the nonzero common factors before solving for $p$

Solution:

Let $p$ be the probability of heads. The condition gives $5p(1-p)^4=10p^2(1-p)^3.$ The stated nonzero condition permits cancellation, yielding $1-p=2p,$ so $p=\frac{1}{3}.$ The probability of exactly three heads is $\binom53\left(\frac13\right)^3\left(\frac23\right)^2=\frac{40}{243}.$ Therefore $i+j=40+243=283.$

1989 AIME Problem 5

Answer: 283

Difficulty rating: 2110

Small Hint:

Big Hint:

Solution:

Problem 5 in Other Years