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

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.

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Hope it helps.
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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

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.

Hope this makes sense

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?

Yes, that's correct.
_________________

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

The probability that Kim will be selected to complete one of the three tasks = P(Kim will be selected for task A) + P(Kim will be selected for task B) + P(Kim will be selected for task C)

Let’s calculate the probability that Kim will be selected for task A:

P(Kim will be selected for task A) = 1/8

Next, we calculate the probability that Kim will be selected for task B (i.e., she won’t be selected for task A):

P(Kim will be selected for task B) = P(Kim won’t be selected for task A) x P(Kim will be selected for task B) = 7/8 x 1/7 = 1/8

Finally, let’s calculate the probability that Kim will be selected for task C (i.e., she won’t be selected for either task A or task B):

P(Kim will be selected for task C) = P(Kim won’t be selected for task A) x P(Kim won’t be selected for task B) x P(Kim will be selected for task C) = 7/8 x 6/7 x 1/6 = 1/8

Thus, the probability that Kim will be selected for one of the three tasks is ⅛ + ⅛ + ⅛ = ⅜.

Alternate Solution:

We will use the fact that

P(Kim will be selected for one of the three tasks) + P(Kim will not be selected for any tasks) = 1

To calculate P(Kim will not be selected for any tasks), we note that this event happens if Kim is not selected for task A, B or C. Since there is a 1/8 probability that Kim is selected for task A, the probability that she is not selected for task A is 1 - 1/8 = 7/8. Similarly, the probability that she is not selected for tasks B and C are 6/7 and 5/6, respectively. Thus, the probability that she is not selected for any tasks is 7/8 x 6/7 x 5/6 = 5/8.

Now, since P(Kim will be selected for one of the three tasks) = 1 - P(Kim will not be selected for any tasks) and since P(Kim will not be selected for any tasks) = 5/8, we conclude that P(Kim will be selected for one of the three tasks) = 1 - 5/8 = 3/8.

Answer: B
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Given: 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.

Asked: What is the probability that Kim will be selected to complete one of the three tasks?

Probability = 7C2 * 3! / 8C3 * 3! = 7*6*3 / 8*7*6 - 3/8

IMO B