A certain class consists of 8 students, including Kim.

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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

The OA is B.

I get the solution as follows,

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 .
Option B.

Can anyone explain another way to solve it, please. Thanks!

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Hi All,

We're told that 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. We're asked for the probability that Kim will be selected to complete one of the three tasks.

With these types of probability questions, there are usually a couple of different ways to get to the correct answer. In many cases, it's fairly easy to calculate what you DON'T want to have happen - and then subtract that fraction from 1 to determine the probability of what you DO want to happen.

The probability of Kim NOT being assigned any of the tasks = (7/8)(6/7)(5/6)

Notice how the most of the numbers in the numerator and denominator of that product 'cancel out', leaving us with 5/8. Thus, the probability that Kim is assigned one of the takes is 1 - 5/8 = 3/8

Final Answer: B

GMAT assassins aren't born, they're made,
Rich