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A certain class consists of 8 students, including Kim. Each [#permalink]

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04 Mar 2013, 10:19

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A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?

Re: A certain class consists of 8 students, including Kim. Each [#permalink]

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04 Mar 2013, 10:41

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Number of ways to select three students to complete the three tasks as described = 8C1 x 7C1 x 6C1 = 8 x 7 x 6

Number of ways to select the three students to complete the three tasks as described such that Kim is selected to do at least one = 7 x 6 + 7 x 6 + 7 x 6 = 7 x 6 x 3 [Here is the logic: If Kim is to do the first task, select the second student in 7C1 ways, and the third in 6C1 ways. If Kim is to do the second task, select a student to the first in 7C1 ways and a student to do the third in 6C1 ways. Similarly for the third task]

Re: A certain class consists of 8 students, including Kim. Each [#permalink]

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04 Mar 2013, 11:46

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megafan wrote:

A certain class consists of 8 students, including Kim. Each day, three tasks must be completed and are assigned as follows: one of the 8 students is selected at random to complete Task A, one of the remaining 7 students is selected at random to complete Task B, and one of the remaining six students is selected at random to complete Task C. What is the probability that Kim will be selected to complete one of the three tasks?

(A) \(\frac{1}{3}\)

(B) \(\frac{3}{8}\)

(C) \(\frac{1}{24}\)

(D) \(\frac{1}{336}\)

(E) \(\frac{1}{512}\)

30-second approach: consider these 8 students in a row. The 3 tasks will be assigned to the first 3 students. Now, what is the probability that Kim will be among the first 3 students? 3/8.

Re: A certain class consists of 8 students, including Kim. Each [#permalink]

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11 Mar 2013, 13:30

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Since the question is asking find the probability of completing one of the three task: Probability of Kim doing task A = 1/8 Probability of Kim doing task B = 1/8 Probability of Kim doing task C = 1/8

So the probability of Kim doing one of the three tasks = 1/8 + 1/8 + 1/8 = 3/8 Answer B

Note: we are not using 1/7 and 1/6 as probability for tasks B and C because it is not specified it has to go in order of A, B and C.

Re: A certain class consists of 8 students, including Kim. Each [#permalink]

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23 May 2013, 21:46

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The question asks: probability that Kim will be selected to complete one of the three tasks which is equal to1-( the probability that kim wil not be selected in any of the three tasks) 1-(7/8x6/7x5/6)=3/8

Is my approach on the problem correct?Can someone please clarify it? _________________

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Re: A certain class consists of 8 students, including Kim. Each [#permalink]

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24 May 2013, 01:20

Expert's post

skamal7 wrote:

The question asks: probability that Kim will be selected to complete one of the three tasks which is equal to1-( the probability that kim wil not be selected in any of the three tasks) 1-(7/8x6/7x5/6)=3/8

Is my approach on the problem correct?Can someone please clarify it?

Re: A certain class consists of 8 students, including Kim. Each [#permalink]

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25 May 2013, 04:43

skamal7 wrote:

The question asks: probability that Kim will be selected to complete one of the three tasks which is equal to1-( the probability that kim wil not be selected in any of the three tasks) 1-(7/8x6/7x5/6)=3/8

Is my approach on the problem correct?Can someone please clarify it?

sure, a smart approach. it can be explained as follows probability of A -1/8 probability of B - non ocurrence of A & Occurrence of B-7/8*1/7=1/8 probability of C 7/8*6/7*1/6=1/8 probability of A or B or C =1/8+1/8+1/8=3/8

Re: A certain class consists of 8 students, including Kim. Each [#permalink]

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21 Jul 2014, 22:54

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Re: A certain class consists of 8 students, including Kim. Each [#permalink]

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28 Aug 2015, 21:25

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