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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 == Problem ==
-When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to <math>0^{}_{}</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij^{}_{}</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3_{}^{}</math> out of <math>5^{}_{}</math> flips. Find <math>i+j^{}_{}</math>.
+When a certain biased coin is flipped five times, the [[probability]] of getting heads exactly once is not equal to <math>0</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>.
 == Solution ==
-{{solution}}
+===Solution 1 ===
+Denote the probability of getting a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>. After canceling out terms, we get <math>1 - h = 2h</math>, so <math>h = \frac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>.
+=== Solution 2 ===
+Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math>. To find the probability getting <math>3</math> heads, we need to find <math>{5\choose3}\dfrac{(h)^3(t)^2}{(h + t)^5} =10\cdot\dfrac{(h)^3(2h)^2}{(h + 2h)^5}</math> (recall that <math>h</math> cannot be <math>0</math>). The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>.
 == See also ==
-* [[1989 AIME Problems/Problem 6|Next Problem]]
+{{AIME box|year=1989|num-b=4|num-a=6}}
-* [[1989 AIME Problems/Problem 4|Previous Problem]]
-* [[1989 AIME Problems]]
+[[Category:Intermediate Combinatorics Problems]]
+{{MAA Notice}}

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