1980 AHSME Problems/Problem 20

Problem

A box contains $2$ pennies, $4$ nickels, and $6$ dimes. Six coins are drawn without replacement, with each coin having an equal probability of being chosen. What is the probability that the value of coins drawn is at least $50$ cents?

$\text{(A)} \ \frac{37}{924} \qquad \text{(B)} \ \frac{91}{924} \qquad \text{(C)} \ \frac{127}{924} \qquad \text{(D)}\ \frac{132}{924}\qquad \text{(E)}\ \text{none of these}$

Solution 1

We want the number of Successful Outcomes over the number of Total Outcomes. We want to calculate the total outcomes first. Since we have $12$ coins and we need to choose $6$ , we have $\binom{12}{6}$ = $924$ Total outcomes. For our successful outcomes, we can have $(1) 1$ penny and $5$ dimes, $2$ nickels and $4$ dimes, $1$ nickel and $5$ dimes, or $6$ dimes.

For the case of $1$ penny and $5$ dimes, there are $\binom{6}{5}$ ways to choose the dimes and $2$ ways to choose the pennies. That is $6 \cdot 2 = 12$ successful outcomes. For the case of $2$ nickels and $4$ dimes, we have $\binom{6}{4}$ ways to choose the dimes and $\binom{4}{2}$ ways to choose the nickels. We have $15 \cdot 6$ = $90$ successful outcomes. For the case of $1$ nickel and $5$ dimes, we have $\binom{4}{1} \cdot \binom{6}{5} = 24$ . Lastly, we have $6$ dimes and $0$ nickels, and $0$ pennies, so we only have one case. Therefore, we have $\dfrac {12 + 90 + 24 + 1}{924} = \dfrac{127}{924}$ = $\boxed{C}$

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1980 AHSME Problems/Problem 20

Problem

Solution 1

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