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megafan
Joined: 28 May 2009
Last visit: 15 Oct 2017
Posts: 137
Given Kudos: 91
Location: United States
Concentration: Strategy, General Management
Schools: Stern PT Fall'18 Booth PT '18
GMAT Date: 03-22-2013
GPA: 3.57
WE:Information Technology (Consulting)
04 Mar 2013, 10:19
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
71% (01:44) correct
29%
(02:12)
wrong
based on 266
sessions
History
Date
Time
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Not Attempted Yet
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) \(\frac{1}{512}\)
(A) \(\frac{1}{3}\)
(B) \(\frac{3}{8}\)
(C) \(\frac{1}{24}\)
(D) \(\frac{1}{336}\)
(E) \(\frac{1}{512}\)
04 Mar 2013, 11:46
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megafan
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.
Answer: B.
Similar questions to practice:
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Hope it helps.
General Discussion
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]
Therefore probability = (7x6x3)/(8x7x6) = 3/8
Option B
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]
Therefore probability = (7x6x3)/(8x7x6) = 3/8
Option B
