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A certain class consists of 8 students, including Don. Each

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

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New post 01 Mar 2011, 04:39
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A
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C
D
E

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A certain class consists of 8 students, including Don. 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 Don will be selected to complete one of the three tasks?

A. 1/3
B. 3/8
C. 1/24
D. 1/336
E. 1/512
[Reveal] Spoiler: OA

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Re: word problem - Probability [#permalink]

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New post 01 Mar 2011, 04:47
\(\frac{1}{8}*1*1 + \frac{7}{8}*\frac{1}{7}*1+\frac{7}{8}*\frac{6}{7}*\frac{1}{6}\)

\(3*\frac{1}{8} = \frac{3}{8}\)

Ans: "B"
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Re: word problem - Probability [#permalink]

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New post 01 Mar 2011, 06:18
tirupatibalaji wrote:
A certain class consists of 8 students, including Don. 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 Don will be selected to complete one of the three tasks?
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512


Alternately you can find the probability that among 3 students chosen to complete the tasks won't be Don and subtract it from 1:
\(P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}\);

The same with probability approach: \(P=1-\frac{7}{8}*\frac{6}{7}*\frac{5}{6}=\frac{3}{8}\).

Answer: B.
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Re: word problem - Probability [#permalink]

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New post 01 Mar 2011, 06:21
P(don will not be selected to complete any) = (1 - 1/8) (1 - 1/7) (1 - 1/6) = 5/8

P(don will be selected atleast once) = 1 - 5/8 = 3/8
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Re: word problem - Probability [#permalink]

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New post 01 Mar 2011, 07:14
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There's no need for a complicated solution here. We're selecting 3 people from 8 and we want to know the probability Don is selected. That's just like lining the 8 people up, and asking the probability Don is one of the first 3 people in line. There are 8 spots for Don, 3 of which are among the first three spots, so the probability must be 3/8.

edit - Or to look at it a different way, each of the 8 people is as likely as anyone else to be selected for A, for B, or for C. That is, the probability Don is selected for A is 1/8, for B is also 1/8 and for C is 1/8, so the probability he is selected for one of the three is 1/8 + 1/8 + 1/8 = 3/8.
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Re: word problem - Probability   [#permalink] 01 Mar 2011, 07:14
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