Consider the following for the two (02) items that follow: The...

Let $E_1$, $E_2$, and $E_3$ be the events that $A$, $B$, and $C$ become managers respectively.

Given probabilities are:
$P(E_1) = \dfrac{3}{10}$
$P(E_2) = \dfrac{1}{2}$
$P(E_3) = \dfrac{4}{5}$

Let $X$ be the event that the bonus scheme is introduced. The conditional probabilities are:
$P(X|E_1) = \dfrac{4}{9}$
$P(X|E_2) = \dfrac{2}{9}$
$P(X|E_3) = \dfrac{1}{3}$

We need to find the probability that $B$ was appointed as the manager given that the bonus scheme was introduced, which is $P(E_2|X)$.

Using Bayes' theorem:
$P(E_2|X) = \dfrac{P(E_2)P(X|E_2)}{P(E_1)P(X|E_1) + P(E_2)P(X|E_2) + P(E_3)P(X|E_3)}$

Calculating the individual terms:
$P(E_1)P(X|E_1) = \dfrac{3}{10} \times \dfrac{4}{9} = \dfrac{12}{90} = \dfrac{6}{45}$

$P(E_2)P(X|E_2) = \dfrac{1}{2} \times \dfrac{2}{9} = \dfrac{1}{9} = \dfrac{5}{45}$

$P(E_3)P(X|E_3) = \dfrac{4}{5} \times \dfrac{1}{3} = \dfrac{4}{15} = \dfrac{12}{45}$

Total probability of introducing the bonus scheme is:
$P(X) = \dfrac{6}{45} + \dfrac{5}{45} + \dfrac{12}{45} = \dfrac{23}{45}$

Substituting the values into Bayes' theorem:
$P(E_2|X) = \dfrac{\dfrac{5}{45}}{\dfrac{23}{45}} = \dfrac{5}{23}$

Answer: $\dfrac{5}{23}$