# SOLUTION: Consider a large bag of coins, consisting of only quarters (\$0.25), dimes (\$0.10), and nickels (\$0.05). 40% of the coins are nickels, 35% are dimes, and 25% are quarters. Random

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: Consider a large bag of coins, consisting of only quarters (\$0.25), dimes (\$0.10), and nickels (\$0.05). 40% of the coins are nickels, 35% are dimes, and 25% are quarters. Random      Log On

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 Question 534682: Consider a large bag of coins, consisting of only quarters (\$0.25), dimes (\$0.10), and nickels (\$0.05). 40% of the coins are nickels, 35% are dimes, and 25% are quarters. Randomly select 5 coins from the bag. What is the probability that your selection is worth at least \$1.00? (Note that the only way to make \$1 with 5 coins is with four (4) quarters and one (1) other coin.)Answer by Edwin McCravy(8896)   (Show Source): You can put this solution on YOUR website!Consider a large bag of coins, consisting of only quarters (\$0.25), dimes (\$0.10), and nickels (\$0.05). 40% of the coins are nickels, 35% are dimes, and 25% are quarters. ``` You must draw either 5 quarters of 4 quarters and a nickel or dime. Suppose there are 100N coins consisting of 40N nickels, 35N dimes, and 25N quarters. Note: there are 75N non-quarters The number of successes equals the number of ways to draw 5 quarters is C(25N,5) plus the number of ways to draw 4 quarters and 1 non-quarter is C(25N,4)·(75N) So the number of successes is C(25N,5) + C(25N,4)·(75N) + Put the (75N) in the numerator of the second fraction + We get a least common denominator of 5! by multiplying the second fraction by + + Factor the numerator: The number of possible ways is the number of ways to select any 5 coins from the 100N coins: C(100N,5) = So the probability is ÷ × × The first fraction is , exactly . If N is large the next three are also very nearly or , and the last fraction is very nearly , or 4. So that is very nearly That's not exact, but it is very close if N is large. Edwin```