If the bonus scheme has been introduced, then what is the probability that the manager appointed was B?

Calculation:

Given,

The probability that A , B , and C  become managers are:

\( P(A) = \frac{3}{10}, \, P(B) = \frac{1}{2}, \, P(C) = \frac{4}{5} \)

The conditional probabilities that the bonus scheme is introduced are:

\( P(D|A) = \frac{4}{9}, \, P(D|B) = \frac{2}{9}, \, P(D|C) = \frac{1}{3} \)

We need to find the probability that B  is the manager given that the bonus scheme has been introduced using Bayes' Theorem:

\( P(B|D) = \frac{P(D|B)P(B)}{P(D)} \)

First, calculate the total probability P(D)  that the bonus scheme is introduced:

\( P(D) = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) \)

Substitute the values:

\( P(D) = \left( \frac{4}{9} \times \frac{3}{10} \right) + \left( \frac{2}{9} \times \frac{1}{2} \right) + \left( \frac{1}{3} \times \frac{4}{5} \right) \)

Simplifying:

\( P(D) = \frac{12}{90} + \frac{2}{18} + \frac{4}{15} \)

Find a common denominator (LCD = 90):

\( P(D) = \frac{12}{90} + \frac{10}{90} + \frac{24}{90} = \frac{46}{90} = \frac{23}{45} \)

Now, use Bayes' Theorem to find P(B|D) :

\( P(B|D) = \frac{P(D|B)P(B)}{P(D)} = \frac{\left( \frac{2}{9} \times \frac{1}{2} \right)}{\frac{23}{45}} \)

Simplifying:

\( P(B|D) = \frac{\frac{2}{18}}{\frac{23}{45}} = \frac{\frac{1}{9}}{\frac{23}{45}} = \frac{5}{23} \)

The probability that the manager appointed was B , given that the bonus scheme has been introduced, is 5/23.

Hence, the correct answer is Option 1.