A company needs to choose a team of 4 persons from a group of 3 content…

Concept: When a team is chosen at random from a pool made up of distinct sub-groups, the probability that it contains an exact number of members from one sub-group follows the classical (hypergeometric) probability rule: P = [C(k, r) x C(n - k, m - r)] / C(n, m), where n is the total pool size, m is the team size, k is the size of the target sub-group, and r is how many members of that sub-group are required.

1. Total people available = 3 content managers + 3 engineers + 5 client engagement managers = 11.

2. Total ways to form any 4-person team from these 11 people = C(11, 4) = (11 x 10 x 9 x 8) / (4 x 3 x 2 x 1) = 330.

3. For exactly 3 client engagement managers on the team, 3 of the 4 seats must come from the 5 client engagement managers, and the remaining 1 seat must come from the other 6 people (3 content managers + 3 engineers).

4. Ways to choose 3 client engagement managers out of 5 = C(5, 3) = (5 x 4 x 3) / (3 x 2 x 1) = 10.

5. Ways to choose the remaining 1 member out of the other 6 people = C(6, 1) = 6.

6. Favourable outcomes = 10 x 6 = 60.

7. Required probability = favourable outcomes / total outcomes = 60 / 330 = 2/11.

Cross-check: summing C(5, r) x C(6, 4 - r) / 330 for r = 0, 1, 2, 3, 4 gives 15/330 + 100/330 + 150/330 + 60/330 + 5/330 = 330/330 = 1, confirming the probabilities across every possible count of client engagement managers on the team are internally consistent, which supports 60/330 = 2/11 as the value for exactly 3.