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probability/combinatorics

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probability/combinatorics

by shahdevine » Tue Sep 01, 2009 4:05 pm
There are 3 dimes and 3 quarters in a coin purse. If 4 coins are chosen at random from the purse, what is the probability of getting more than 65 cents?

a)1/5
b)2/5
c)1/2
d)3/5
e)4/5

OA after some discussion

65 cents requires some combination of QQQQ, QQQD,QQDD, but I get stuck with these probability PC problems. Could someone breakdown clear, logical steps?

thx
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by PussInBoots » Tue Sep 01, 2009 11:55 pm
(3C2*3C2 + 3C1 * 3C3) / 6C4
(2 dimes * 2 quarter + 1 dime * 3 quarters) / pick 4 coins out of 6

Answer is E
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by tom4lax » Wed Sep 02, 2009 10:29 am
My way of thinking about it:

Pick 1 Dime and 3 Quarters (85c)
3!/1!2! * 3!/3! = 3

Pick 2 Dimes and 2 Quarters (70c)
3!/2!1! * 3!/2!1! = 9

Pick 3 Dimes and 1 Quarter (55c)
Same as above = 3

total over 65c / total possibilities
12/15 = 4/5 = E
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