Question

The probabilities of x, y and z becoming manager are$\frac{4}{9},\frac{2}{9}\&\frac{1}{3}$ respectively. The probabilities that the bonus scheme will be introduced if x,y and z become managers are$\frac{3}{10},\frac{1}{2}\&\frac{2}{5}$ respectively. Find (i) What is the probatility that the bonus scheme will be introduced? (ii) if the bonus scheme has teen introduced, What is the probability that the manager appointed wax x?

Answer 1

$P\left(x\right)=\frac{4}{9},P\left(y\right)=\frac{2}{9},P\left(z\right)=\frac{1}{3}$

Let event $A$ be the bonus scheme introduced

so, $P\left(A/x\right)=\frac{3}{10}$ and $P\left(A/y\right)=\frac{1}{2}$

$\left(1\right)$ $P\left(A\right)=P\left(x\right).P\left(a/x\right)+P\left(y\right).P\left(A/y\right)+P\left(z\right).P\left(A/z\right)$

$=\frac{4}{9}\times \frac{3}{10}+|cfrac29\times \frac{1}{2}+\frac{1}{2}\times \frac{2}{5}$

$=\frac{6+5+6}{45}=\frac{17}{45}$

$\left(2\right)$ Probability that manager appointed if bonus introduced

$P\left(z/A\right)=\frac{P\left(A/z\right).P\left(z\right)}{P\left(A\right)}$

$=\frac{2\times 45}{15\times 17}=\frac{6}{17}$