Question

For A, B, and C, the chances of being selected as the manager of a firm are 4 : 1 : 2 respectively. The probabilities for them to introduce a radical change in the marketing strategy are 0.3, 0.8 and 0.5 respectively. If a change takes place; find the probability that it is due to the appointment of B.

Answer 1

Let $E_1,E_2,E_3$ and E be the events as defined below:
$E_1$ = A is selected as Manager
$E_2$ = B is selected as Manager
$E_3$ = C is selected as Manager
and E = radical change occurs in marketing strategy
$P\left(E_1\right)=\frac{4}{4+1+2}=\frac{4}{7}$

$P\left(E_2\right)=\frac{1}{4+1+2}=\frac{1}{7}$

$P\left(E_3\right)=\frac{2}{4+1+2}=\frac{2}{7}$

Given, $P\left(\frac{E}{E_1}\right)=0.3$

$P\left(\frac{E}{E_2}\right)=0.8$

$P\left(\frac{E}{E_3}\right)=0.5$

We want to find the probability that the radical change in marketing strategy occured due to the appointment of B i.e., we want to find $P\left(E_2/E\right)$

By Baye's theorem,

$P\left(\frac{E_2}{E}\right)=\frac{P\left(E_2\right).P\left(\frac{E}{E_2}\right)}{P\left(E_1\right)P\left(\frac{E}{E_1}\right)+P\left(E_2\right).P\left(\frac{E}{E_2}\right)+P\left(E_3\right)P\left(\frac{E}{E_3}\right)}$

$=\frac{\frac{1}{7}\times 0.8}{\frac{4}{7}\times 0.3+\frac{1}{7}\times 0.8+\frac{2}{7}\times 0.5}$

$=\frac{0.8}{1.2+0.8+1}$

$=\frac{0.8}{3}=\frac{8}{30}=\frac{4}{15}$