SOLUTION: A bag contains 3 nickels, 4 dimes, and 5 quarters. If two coins are selected at random without replacing the first one selected, what is the probability that the two coins match?
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-> SOLUTION: A bag contains 3 nickels, 4 dimes, and 5 quarters. If two coins are selected at random without replacing the first one selected, what is the probability that the two coins match?
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Question 297867: A bag contains 3 nickels, 4 dimes, and 5 quarters. If two coins are selected at random without replacing the first one selected, what is the probability that the two coins match? Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! 3 nickels
4 dimes
5 quarters
12 total coins
P(N)=3/12
P(D)=4/12
P(Q)=5/12
Once one coin is chosen then the total number of coins drops to 11 and the probability of each coin is adjusted.
P(N if nickel is chosen first)=2/11
P(D if dime is chosen first)=3/11
P(Q if quarter is chosen first)=4/11
Since each event is independent, multiply the probabilities.
P(nickel then nickel)=(3/12)(2/11)=1/22
P(dime then dime)=(4/12)(3/11)=1/11
P(quarter then quarter)=(5/12)(4/11)=5/33
