Question

A committee will be formed with 4 managers and 3 engineers selected randomly without replacement from 11 managers and 16 engineers.

What is the conditional probability that engineer Al is on the committee given that engineer Jane is on the committee?
Round your answer to three decimal places (e.g. 98.765).

Would the answer be ((3/16)x(2/15))/(3/16) ---> (2/15) ----> .1333333

Answer 1

Conditional probability that engineer Al is on the committee given that engineer Jane is on the committee

= Probability that engineer Al and Jane is on the committee / Probability that engineer Jane is on the committee

Now,

Number of ways to select 4 managers and 3 engineers from 11 managers and 16 engineers = ¹¹C₄ * ¹⁶C₃

If Jane (engineer) is on the committee, number of ways to select 4 managers and 2 engineers from 11 managers and 15 engineers = ¹¹C₄ * ¹⁵C₂

Probability that engineer Jane is on the committee =  ¹¹C₄ * ¹⁵C₂ / ¹¹C₄ * ¹⁶C₃

=  ¹⁵C₂ /  ¹⁶C₃ = (15 * 14 /2 * 1) / (16 * 15 * 14 / 3 * 2 * 1)

= 3/16

If Jane (engineer) and Al (engineer) is on the committee, number of ways to select 4 managers and 1 engineer from 11 managers and 14 engineers = ¹¹C₄ * ¹⁴C₁

Probability that engineer Jane and Al is on the committee =  ¹¹C₄ * ¹⁴C₁/ ¹¹C₄ * ¹⁶C₃

=  ¹⁴C₁ /  ¹⁶C₃ = (14 / 1) / (16 * 15 * 14 / 3 * 2 * 1)

= 6/240 = 1/40

Conditional probability that engineer Al is on the committee given that engineer Jane is on the committee

= Probability that engineer Al and Jane is on the committee / Probability that engineer Jane is on the committee

= (1/40) / (3/16)

= 2/15

= 0.1333333