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# Ten students, including Jack and Jil, are each entered into a race

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Ten students, including Jack and Jil, are each entered into a race  [#permalink]

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17 Dec 2018, 19:43
8
00:00

Difficulty:

75% (hard)

Question Stats:

53% (02:52) correct 47% (02:25) wrong based on 59 sessions

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Ten students, including Jack and Jil, are each entered into a race to win one of five bicycles. If each student holds one entry in the race, what is the probability that Jack will win a bicycle but Jil will not?

A) 4/35
B) 4/15
C) 2/5
D) 49/105
E) 5/18
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Re: Ten students, including Jack and Jil, are each entered into a race  [#permalink]

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17 Dec 2018, 23:16
1
ruchik wrote:
Ten students, including Jack and Jil, are each entered into a race to win one of five bicycles. If each student holds one entry in the race, what is the probability that Jack will win a bicycle but Jil will not?

A) 4/35
B) 4/15
C) 2/5
D) 49/105
E) 5/18

To win a bike, Jack should be in the top 5 and to not win a bike, Jill should be in the bottom 5.

For Jack, select a place in top 5 in 5 ways (1st/2nd/3rd/4th/5th)
For Jill, select a place in bottom 5 in 5 ways (6th/7th/8th/9th/10th)

For total cases, Jack has 10 spots available and Jill has 9 spots available.

Required Probability = 5*5/10*9 = 5/18

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Ten students, including Jack and Jil, are each entered into a race  [#permalink]

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Updated on: 26 Dec 2018, 23:05
P(Jack wins but Jill does not) = Total number of favorable outcomes / Total possible outcomes.

Total favorable outcomes = 5 x 5 (Jack picks any top 5 position while Jill picks any bottom 5) = 25.
Total number of possible outcomes = 10 x 9 (Jack picks 1 out of 10 possible slots, then Jill picks 1 out of the remaining 9) = 90.
P(Jack wins but Jill does not) = 25/90 = 5/18 (Option E).

Another way to do this is:
Total number of ways to select 5 out of 10 people to win 5 bicycles = 10C5
Total number of ways to select Jack but not Jill for 5 people = 8C4 (Jack and Jill are excluded from the pool; Jack must be selected, so we only need to pick 4 random persons out of the pool of 8 remaining candidates)
P(Jack wins but Jill does not) = 8C4 / 10C5 = 5/18 (Option E).

Originally posted by tinytiger on 18 Dec 2018, 08:53.
Last edited by tinytiger on 26 Dec 2018, 23:05, edited 1 time in total.
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Re: Ten students, including Jack and Jil, are each entered into a race  [#permalink]

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26 Dec 2018, 12:07
tinytiger wrote:
P(Jack wins but Jill does not) = Total number of favorable outcomes / Total possible outcomes.

Total number of ways to select Jack but not Jill for 5 people = 8C3 (Jack and Jill are excluded from the pool; Jack must be selected, so we only need to pick 3 random persons out of the pool of 8 remaining candidates)
P(Jack wins but Jill does not) = 8C3 / 10C5 = 5/18 (Option E).

Why do you pick 3 and not 4 if it is a group of 5??
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Re: Ten students, including Jack and Jil, are each entered into a race  [#permalink]

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26 Dec 2018, 18:29
tinytiger wrote:
P(Jack wins but Jill does not) = Total number of favorable outcomes / Total possible outcomes.

Total number of ways to select Jack but not Jill for 5 people = 8C3 (Jack and Jill are excluded from the pool; Jack must be selected, so we only need to pick 3 random persons out of the pool of 8 remaining candidates)
P(Jack wins but Jill does not) = 8C3 / 10C5 = 5/18 (Option E).

Why do you pick 3 and not 4 if it is a group of 5??

I think this might just be a typo. 8c3 / 10c5 doesn't come out to 5/18, but 8c4 / 10c5 does. That also makes sense given what's in the problem: you need to choose FOUR other people, aside from Jack, who will win bikes.
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Re: Ten students, including Jack and Jil, are each entered into a race  [#permalink]

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26 Dec 2018, 23:08
Thanks for the spot, typo corrected for in my original post. Correct answer is to pick FOUR, not three (am avoiding the usage of numerals since I may have typed too quickly in my previous post).
Re: Ten students, including Jack and Jil, are each entered into a race   [#permalink] 26 Dec 2018, 23:08
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