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Difference between revisions of "1983 AHSME Problems/Problem 15"

m (Fixed problem statement and formatting)
(Further fixed problem statement and improved clarity of solution)
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==Problem==
 
==Problem==
  
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Three balls marked <math>1,2</math> and <math>3</math>, are placed in an urn. One ball is drawn, its number is recorded, and then the ball is returned to the urn. This process is repeated and then repeated once more, and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is <math>6</math>, what is the probability that the ball numbered <math>2</math> was drawn all three times?  
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Three balls marked <math>1,2</math> and <math>3</math> are placed in an urn. One ball is drawn, its number is recorded, and then the ball is returned to the urn. This process is repeated and then repeated once more, and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is <math>6</math>, what is the probability that the ball numbered <math>2</math> was drawn all three times?  
  
 
<math>
 
<math>
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==Solution==
 
==Solution==
  
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Since a ball is drawn three times and the sum is <math>6</math>, the only possibilities of this happening are permutations of <math>1, 2, 3</math> and <math>2, 2, 2</math>. Therefore, there are 7 possibilities so the probability that the ball numbered <math>2</math> was drawn all three times would be <math>\boxed{\textbf{(C)}\ \frac{1}{7}}</math>.
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Since a ball is drawn three times and the sum is <math>6</math>, the only ways this can happen are the permutations of <math>1, 2, 3</math> and <math>2, 2, 2</math>. Therefore, with there being <math>3! + 1 = 7</math> equally-likely possibilities in total, the probability that the ball numbered <math>2</math> was drawn all three times is <math>\boxed{\textbf{(C)}\ \frac{1}{7}}</math>.

Revision as of 19:56, 26 January 2019

Problem

Three balls marked $1,2$ and $3$ are placed in an urn. One ball is drawn, its number is recorded, and then the ball is returned to the urn. This process is repeated and then repeated once more, and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is $6$, what is the probability that the ball numbered $2$ was drawn all three times?

$\textbf{(A)} \ \frac{1}{27} \qquad \textbf{(B)} \ \frac{1}{8} \qquad \textbf{(C)} \ \frac{1}{7} \qquad \textbf{(D)} \ \frac{1}{6} \qquad \textbf{(E)}\ \frac{1}{3}$

Solution

Since a ball is drawn three times and the sum is $6$, the only ways this can happen are the permutations of $1, 2, 3$ and $2, 2, 2$. Therefore, with there being $3! + 1 = 7$ equally-likely possibilities in total, the probability that the ball numbered $2$ was drawn all three times is $\boxed{\textbf{(C)}\ \frac{1}{7}}$.

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