Question

A is targeting to B, B and C are targeting to A.The probability of hitting the target by A, B and Care $\frac{2}{3},\frac{1}{2},\frac{1}{3}$ respectively. Show that the probability that B hits the target and C does not is $\frac{1}{2}$.

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $\frac{1}{2}$

Probability of A being hit = $P\left(E\right)=1-\left(ProbabilityofBmissing\right)\ast \left(ProbabilityofCmissing\right)=1-(1-\frac{1}{2})\times (1-\frac{1}{3})=\frac{2}{3}$
Probability that A is hit by B but not C = $P\left(\right(B\cap \overline{C}\left)|E\right)=\frac{P\left(B\right)\ast P\left(\overline{C}\right)}{P\left(E\right)}=\frac{\frac{1}{2}\ast \frac{2}{3}}{\frac{2}{3}}=\frac{1}{2}$