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GOODMAN
 3 years ago
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
GOODMAN
 3 years ago
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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KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2You 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)=\[{8 \over 25} \cdot {4 \over 24} + {4 \over 25}\cdot{8 \over 24}\]

GOODMAN
 3 years ago
Best ResponseYou've already chosen the best response.0See, i tried all that, but i ended up being wrong :(

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2For 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2So the answer I gave doesn't work?

GOODMAN
 3 years ago
Best ResponseYou've already chosen the best response.0No, KingGeorge, i think you are correct, because i did it Luis's way, and i got it incorrect.

GOODMAN
 3 years ago
Best ResponseYou've already chosen the best response.0I understand A. But im a little confused on B.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2An 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.2A 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.

GOODMAN
 3 years ago
Best ResponseYou've already chosen the best response.0About the first way you explained it, why did you add them together? Dont you just get one of the products?

GOODMAN
 3 years ago
Best ResponseYou've already chosen the best response.0Ohh, 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?

GOODMAN
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, i see now, thanks a ton!!
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