Difference between revisions of "1989 AIME Problems/Problem 5"

Latest revision as of 23:07, 24 June 2023

Problem

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 $\frac ij$ , in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$ .

Solution

Solution 1

Denote the probability of getting a heads in one flip of the biased coin as $h$ . Based upon the problem, note that ${5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3$ . After canceling out terms, we get $1 - h = 2h$ , so $h = \frac{1}{3}$ . The answer we are looking for is ${5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}$ , so $i+j=40+243=\boxed{283}$ .

Solution 2

Denote the probability of getting a heads in one flip of the biased coins as $h$ and the probability of getting a tails as $t$ . Based upon the problem, note that ${5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3$ . After cancelling out terms, we end up with $t = 2h$ . To find the probability getting $3$ heads, we need to find ${5\choose3}\dfrac{(h)^3(t)^2}{(h + t)^5} =10\cdot\dfrac{(h)^3(2h)^2}{(h + 2h)^5}$ (recall that $h$ cannot be $0$ ). The result after simplifying is $\frac{40}{243}$ , so $i + j = 40 + 243 = \boxed{283}$ .

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 [[Category:Intermediate Combinatorics Problems]]
 {{MAA Notice}}
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1989 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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