Question

Three person a b and c aim a target. The probability of their hitting the target are respectively 2/3, 1/4, 1/2. Find the probablity

Answer 1

Probability that one of them hits the target

$P_{a} = \frac{2}{3} *(1-\frac{1}{4})*(1-\frac{1}{2})=\frac{2}{3} *\frac{3}{4}*\frac{1}{2}=\frac{1}{4}$

$P_{b} = \frac{1}{4} *(1-\frac{2}{3})*(1-\frac{1}{2})=\frac{1}{4} *\frac{1}{3}*\frac{1}{2}=\frac{1}{24}$

$P_{c} = (1-\frac{2}{3} )*(1-\frac{1}{4})*\frac{1}{2}=\frac{1}{3} *\frac{3}{4}*\frac{1}{2}=\frac{1}{8}$

$P_{a}+P_{b}+P_{c}=\frac{1}{4}+\frac{1}{24}+\frac{1}{8}=\frac{5}{12}$

Probability that it was the first man

$P = \frac{P_a}{P_{a}+P_{b}+P_{c}}=\dfrac{\frac{1}{4}}{\frac{5}{12}}= \frac{3}{5}$