1974 AHSME Problems/Problem 24

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Problem

A fair die is rolled six times. The probability of rolling at least a five at least five times is

$\mathrm{(A)\ } \frac{13}{729} \qquad \mathrm{(B) \ }\frac{12}{729} \qquad \mathrm{(C) \  } \frac{2}{729} \qquad \mathrm{(D) \  } \frac{3}{729} \qquad \mathrm{(E) \  }\text{none of these}$

Solution

The probability of rolling at least a five on any one roll is $\frac{2}{6}=\frac{1}{3}$. If there are exactly $5$ fives or sixes rolled, there are $\binom{6}{5}=6$ ways to pick which of the rolls are the fives and sixes, and so the probability in this case is $6\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)=\frac{12}{729}$. If there are exactly $6$ fives or sixes rolled, then there is only one way to pick which of the rolls are fives and sixes, so the probability in this case is $\left(\frac{1}{3}\right)^6=\frac{1}{729}$.

Therefore, the total probability is $\frac{12}{729}+\frac{1}{729}=\frac{13}{729}, \boxed{\text{A}}$.

See Also

1974 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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