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In United States currency, a nickel ....

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In United States currency, a nickel ....

by factor26 » Sun Oct 30, 2011 2:05 pm
In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?

a - 8/25

b - 12/35

c - 13/35

d - 9/25

e - 17/25

answer is b please explain how to solve ... thanks!
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by neelgandham » Sun Oct 30, 2011 2:59 pm
factor26 wrote:In United States currency, a nickel is worth 5 cents, a penny is worth 1 cent, and a dime is worth 10 cents. 100 cents equals one dollar. If a hand purse contains 6 nickels, 5 pennies and 4 dimes, what is the probability of picking a coin other than a nickel twice in a row if the first coin picked is not put back?

a - 8/25

b - 12/35

c - 13/35

d - 9/25

e - 17/25

answer is b please explain how to solve ... thanks!
probability of picking a coin other than a nickel (first turn) =9/15
probability of picking a coin other than a nickel (first turn) =8/14

P = (9/15)*(8/14) =(3/5)*(4/7) = [spoiler]12/35[/spoiler]
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by shankar.ashwin » Sun Oct 30, 2011 4:10 pm
6N,5P and 4D. Total - 15. We are interested only in the 9 coins here (excluding the 6N)

So, (9/15) * (8/14) = 12/35 B
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by factor26 » Mon Oct 31, 2011 1:53 pm
I misunderstood the question that was asked. For some reason I though they were looking for the probability of picking either a dime or a penny consectively .... Prob of picking two dimes d x d + prob of two pennies in a row p x p whoops!!! Thanks guys!!!
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