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My thought is to find prob that the second coin is a penny and then calculate 1-that probability since any coin other than the penny is worth at least 5 cents.

If I know the first coin was a dime and not replaced, my reasoning was this is now the same as asking the prob of drawing a penny from the jar with 1 less dime ($74$ total coins). Thus, the prob of drawing a penny would be $24/74=.32$. So, now the prob of the second coin NOT being a penny would be $25/37$ which is not a choice. Why doesn't this thought process work?

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    Just by inspection...all the given answers are much too low. The answer is obviously greater than $\frac 12$, say. – 2017-02-25
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    I'll bet answer $c.$ is a typo...should read $\frac {25}{37}$, as you say. – 2017-02-25
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    thank you for taking the time to answer. – 2017-02-25
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    The problem is simpler than it seems. "given that the first coin is a dime" means you can simply delete one of the dimes (leaving 14 dimes). Then the probability calculation is trivial. – 2017-02-25
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    I agree, typo. 25/37 – 2017-02-25

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Given that a dime was removed from the jar, the jar now has 24 pennies, 17 nickels, 14 dimes, 19 quarters.

The probability that a coin drawn from the jar has value at least 5 cents is 1 minus the probability that a coin drawn from the jar has value less than 5 cents (a penny).

$$ P = 1 - \frac{24}{74} = \frac{50}{74} = \frac{25}{37} $$