Question

team of 4 members is chosen from a team of 3 trainees, 3 engineers and 5 managers. Find the probability that exactly 3 of them are managers?

Answer 1

This is a multinomial distribution with the the number of the members chosen as n = 4

The variables are :

X₁, x₂, x₃

Let :

x₁ = trainees

x₂= Engineers

x₃= managers

The formula for solving this is as follows :

n! / x₁! x₂! x₃! × p₁ˣ¹ × P₂ˣ² × P₃ˣ³

P₁ = 3/11 (probability of selecting a trainee)

P₂ = 3/11( probability of selecting an engineer)

P₃ = 5/11 (probability of selecting a manager)

We want exactly 3 to be managers. This means :

P(x₁ = 1, x₂ = 0, x₃ = 3) + P(x₁ = 0, x₂ = 1, x₃ = 3)

Substituting this we have :

{4!/1!0! 3! × (3/11)¹ × (3/11)⁰ × (5/11)³} + {4!/0!1! 3! × (3/11)⁰ × (3/11)¹ × (5/11)³

4 × 3/11 × (5/11)³ + 4 × 3/11 ×(5/11)³

1500/14641 + 1500/14641

0.10245 + 0.10245 = 0.2049