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$3$ nickels, $1$ dime and $2$ quarters are in a purse. In picking three coins at one time (without replacement), what is the probability of getting a total of at least $35$ cents?

I know that the total ways of selecting 3 coins out of 6 is ${6\choose 3}=20$. In order to have a total at least 35 cents, there must be at least 1 quarter among the 3 chosen coins. But I don't know how to proceed.

One way to do this problem is to find the probability that 3 chosen coins are not quarters. This is ${4\choose 3}=4$ (numbers of ways choosing three coins out of 4 non-quarters). Then I take the total 20 and subtract 4 to get 16. The probability of getting a sum at least 35 is therefore $\frac{16}{20}$.

Is there any other methods to solve this problem?

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Another way would be to just directly calculate the probability of picking 3 non-quarters, and subtract that probability from 1, since at least one coin needs to be a quarter for 35 cents or more:

first draw, 4/6
second draw, 3/5
third draw, 2/4

Multiplying these together:
(4*3*2)/(4*5*6) = 1/5, and 1 - 1/5 = 4/5 = 16/20.

I do think that recognizing that this question can be rephrased as "what's the probability of drawing at least 1 quarter?" is the key to making this problem easier, doing either your way or mine above; there are likely other ways as well though.

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  • $\begingroup$ Just to state explicitly, you will always get at least 35 cents if you have at least 1 quarter, and you will never get at least 35 cents if you have less than 1 quarter. This is the reason the problem splits into only two possibilities. If there was, for example, 1 penny, then you could draw 1 quarter, 1 penny, 1 nickel, and not reach 35 cents. In that case, the problem would not simplify in this way. $\endgroup$ – AlexanderJ93 Oct 15 '18 at 19:36
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    $\begingroup$ Right, I should've mentioned that in the initial answer, I didn't elaborate more at the end. Without that simplification the problem is certainly more challenging. $\endgroup$ – Paul Oct 15 '18 at 19:45

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