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

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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>.
 
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>.
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==See Also==
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{{AHSME box|year=1983|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 23:54, 19 February 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 $\boxed{\textbf{(C)}\ \frac{1}{7}}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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