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Difference between revisions of "1988 AHSME Problems/Problem 28"

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We have <math>w = {5\choose3}p^{3}(1-p)^{2} = 10p^{3}(1-p)^{2}</math>, so we need to solve <math>10p^{3}(1-p)^{2} = \frac{144}{625} \implies p^{3}(1-p)^{2} = \frac{72}{3125}</math>. Now observe that when <math>p=0</math>, the left-hand side evaluates to <math>0</math>, which is less than <math>\frac{72}{3125}</math>; when <math>p=\frac{1}{2}</math>, it evaluates to <math>\frac{1}{32}</math>, which is more than <math>\frac{72}{3125}</math>, and when <math>p=1</math>, it evaluates to <math>0</math> again. Thus, since <math>p^{3}(1-p)^{2}</math> is continuous, the Intermediate Value Theorem tells us there is a solution between <math>0</math> and <math>\frac{1}{2}</math> and another solution between <math>\frac{1}{2}</math> and <math>1</math>, meaning that there is not a unique value of <math>p</math>, so the answer is <math>\boxed{\text{D}}</math>.
  
  

Latest revision as of 13:50, 27 February 2018

Problem

An unfair coin has probability $p$ of coming up heads on a single toss. Let $w$ be the probability that, in $5$ independent toss of this coin, heads come up exactly $3$ times. If $w = 144 / 625$, then

$\textbf{(A)}\ p\text{ must be }\tfrac{2}{5}\qquad \textbf{(B)}\ p\text{ must be }\tfrac{3}{5}\qquad\\ \textbf{(C)}\ p\text{ must be greater than }\tfrac{3}{5}\qquad \textbf{(D)}\ p\text{ is not uniquely determined}\qquad\\ \textbf{(E)}\ \text{there is no value of } p \text{ for which }w =\tfrac{144}{625}$


Solution

We have $w = {5\choose3}p^{3}(1-p)^{2} = 10p^{3}(1-p)^{2}$, so we need to solve $10p^{3}(1-p)^{2} = \frac{144}{625} \implies p^{3}(1-p)^{2} = \frac{72}{3125}$. Now observe that when $p=0$, the left-hand side evaluates to $0$, which is less than $\frac{72}{3125}$; when $p=\frac{1}{2}$, it evaluates to $\frac{1}{32}$, which is more than $\frac{72}{3125}$, and when $p=1$, it evaluates to $0$ again. Thus, since $p^{3}(1-p)^{2}$ is continuous, the Intermediate Value Theorem tells us there is a solution between $0$ and $\frac{1}{2}$ and another solution between $\frac{1}{2}$ and $1$, meaning that there is not a unique value of $p$, so the answer is $\boxed{\text{D}}$.


See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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