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Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5

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Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5  [#permalink]

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New post 24 Mar 2019, 23:57
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Question Stats:

42% (03:08) correct 58% (03:13) wrong based on 19 sessions

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Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let p be the probability that all four slips bear the same number. Let q be the probability that two of the slips bear a number a and the other two bear a number \(b \neq a\). What is the value of q/p?

(A) 162
(B) 180
(C) 324
(D) 360
(E) 720

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Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5  [#permalink]

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New post 25 Mar 2019, 06:14
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Bunuel wrote:
Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let p be the probability that all four slips bear the same number. Let q be the probability that two of the slips bear a number a and the other two bear a number \(b \neq a\). What is the value of q/p?

(A) 162
(B) 180
(C) 324
(D) 360
(E) 720



Since, we have probability out of the same set, and we have probability in both numerator and denominator, the q/p can be taken as (ways of picking two different pairs of number)/(ways of picking all four same). This will save some calculations..

Ways to choose all four - 10 *\(\frac{4!}{4!}\), as all the numbers are similar, so 10 ways
Ways to choose two different pair - 10C2*4C2*4C2, that is (choose two out of 10), then there are 4 of each of these number, so 4C2 ways for each number)

Thus \(\frac{q}{p}=\frac{10C2*4C2*4C2}{10}=\frac{10*9*4*3*4*3}{2*2*2*10}=9*9*2=162\)
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Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5   [#permalink] 25 Mar 2019, 06:14
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Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5

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