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.

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

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

Using the law of total probability, the probability that the bonus scheme will be introduced is:
$P(S) = P(E_1)P(S|E_1) + P(E_2)P(S|E_2) + P(E_3)P(S|E_3)$

Substituting the given values:
$P(S) =$\left(\dfrac{3}{10} \times \dfrac{4}{9}\right) + $\left(\dfrac{1}{2} \times \dfrac{2}{9}\right) + \left(\dfrac{4}{5} \times \dfrac{1}{3}\right)<br><br>$P(S) = $\dfrac{12}{90} + \dfrac{2}{18} + \dfrac{4}{15}<br><br>$ P(S) = $\dfrac{2}{15} + \dfrac{1}{9} + \dfrac{4}{15}<br><br>$ P(S) = $\dfrac{6}{15} + \dfrac{1}{9}<br><br>$ P(S) = $\dfrac{2}{5} + \dfrac{1}{9}<br><br>$ P(S) = $\dfrac{18 + 5}{45} = \dfrac{23}{45}$$<br><br>Answer:$$\dfrac{23}{45}$