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!
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!
