Question

Three persons A, B and C aim a target. The probabilities of their hitting the target are respectively 2/3, 1/4, 1/2. What is the probability that the target will be hit?

• A. 1/8
• B. 3/8
• C. 5/8
• D. 7/8

Answer 1

The correct option is D 7/8

$Let{E}_1.E_2,E_{3}be\text{the events that A,B,C hits the target,}$
$\text{Then}\text{P}\left(E_1\right)=\frac{2}{3},\text{P}\left(E_2\right)=\frac{1}{4},\text{P}\left(E_3\right)=\frac{1}{2}$
$T\text{he target is hit}\text{in following ways:}$
$\text{i)A hit the target and B,C did not hit}$
$\text{ii)B}\text{hit the target and A, C did not hit}$
$\text{iii)C hit the target and B,C did not hit}$
$\text{iv)Aand B hit the target and C did not hit}$
$\text{v)A and C hit the target and B did not hit}$
$\text{vi) B and C hit the target and A did not hit}$
$\text{vii) A,B,C hit the target}$
$\text{So required probability=}\frac{2}{3}\times \frac{3}{4}\times \frac{1}{2}+\frac{1}{4}\times \frac{1}{3}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{3}\times \frac{3}{4}+$
$\frac{2}{3}\times \frac{1}{4}\times \frac{1}{2}+\frac{2}{3}\times \frac{1}{2}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{2}\times \frac{1}{3}+\frac{2}{3}\times \frac{1}{4}\times \frac{1}{2}$
$=\frac{6}{24}+\frac{1}{24}+\frac{3}{24}+\frac{2}{24}+\frac{6}{24}+\frac{1}{24}+\frac{2}{24}$
$=\frac{21}{24}=\frac{7}{8}$