Question 100533: There are 5 nickels, 3 dimes, and 2 quarters. What is the probability of selecting 1 nickel, 1 dime, and 1 quarter in any order with replacement?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! There are 5 nickels, 3 dimes, and 2 quarters. What is the probability of selecting 1 nickel, 1 dime, and 1 quarter in any order with replacement?
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"With replacement" makes all your selections independent.
P(nickel)=5/10= 1/2
P(dime)=3/10
P(quarter)=2/10= 1/5
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So, Prob(nickel dime quarter in any order) = 3!(30/1000) = 6(3/100)= 0.18
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Cheers,
Stan H.
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