A bank contains 3 pennies, 8nickels, 4 dimes, and 10 quarters. Two coins are selected at random. Find the probability of each selection: A) P(2 pennies) B) P(1 nickel and 1 dime) Please explain how you did it, thanks! :D

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You have 3+8+4+10=25 coins total. Thus, P(2 pennies) = \[{3 \over 25}\cdot {2 \over 24}\]And P(1 nickel and 1 dime)=\[{8 \over 25} \cdot {4 \over 24} + {4 \over 25}\cdot{8 \over 24}\]

See, i tried all that, but i ended up being wrong :(

For A, I got it by multiplying the probability of getting a penny on the first try, and on the second try. For B, it's the probability of getting a nickel on the first try times the probability of getting a dime on the second try added to the probability of getting a dime the first try, and a nickel the second try.

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