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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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

Mathematics
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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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So the answer I gave doesn't work?
No, KingGeorge, i think you are correct, because i did it Luis's way, and i got it incorrect.
I understand A. But im a little confused on B.
An alternative way to look at this problem, is that both coins are selected simultaneously, and it's asking for the probability that we got those combinations. In that case, the answers should be as follows A: \[\Large {3\over \binom{25}{2}}\] Since you have 3 ways of choosing 2 pennies from 3, divided by the total number of ways to choose 2 coins. B.\[\Large {{8\cdot4} \over \binom{25}{2}}\]Sicne you have 8 ways of choosing one nickel, and 4 ways of choosing 1 dime. Then you divide by total number of possibilities.
A better explanation of my first solution for B. would be as such. My first method of solving requires that you pick one coin first, and then the second one. This is why I'm adding two terms. The first term is the probability that we chose the nickel first and then the dime, and this is different from the probability that we chose the dime first and then the nickel. Since these are distinct probabilities, we have to add them together.
About the first way you explained it, why did you add them together? Dont you just get one of the products?
Ohh, because in combinations, order doesnt matter, so it could be either way, a nickel then dime or a dime and nickel. Is that correct and is that why we have to add them together?
Exactly.
Okay, i see now, thanks a ton!!
no problem

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