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metallicafan
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17 Feb 2012, 07:50
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
77% (01:30) correct
23%
(02:02)
wrong
based on 1228
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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?
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
Source: Jeff Sackman questions - https://www.gmathacks.com
(A) 1/3
(B) 3/8
(C) 1/24
(D) 1/336
(E) 1/512
In solved in this way:
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
Probability of completing Task A:
\(\frac{1}{8} * \frac{7}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task B:
\(\frac{7}{8} * \frac{1}{7} * \frac{6}{6} = \frac{1}{8}\)
Probability of completing Task C:
\(\frac{7}{8} * \frac{6}{7} * \frac{1}{6} = \frac{1}{8}\)
Therefore: \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = [m]\frac{3}{8}\)[/m]
Is there another way to solve it faster?
17 Feb 2012, 08:23
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metallicafan
The wording of the question makes it harder then it is. The question basically asks: what is the probability that Kim is among the first 3 students out of 8, which is 3/8.
One can also do: \(P=\frac{C^1_1*C^2_7}{C^3_8}=\frac{3}{8}\), where \(C^1_1\) is # of ways to select Kim, \(C^2_7\) is # of ways to select any 2 students out of 7 left and \(C^3_8\) is total # of ways to select 3 students from 8;
Or: you can find the probability that among 3 students selected to complete the tasks there won't be Kim and subtract it from 1:
\(P=1-\frac{C^3_7}{C^3_8}=\frac{3}{8}\);
Or: the same with probability approach: \(P=1-\frac{7}{8}*\frac{6}{7}*\frac{5}{6}=\frac{3}{8}\).
As you can see first approach is easiest and shortest one.
P.S. What about the answer choices?
17 Feb 2012, 07:55
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metallicafan
Since there are total of 8 students and we are selecting 3 of them, then the probability that Kim will be selected is simply 3/8 (she has 3 chances out of 8).
P.S. Please provide answer choices if available.
