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$A$ is targeting to $B,B$ and $C$ are targeting to $A.$ probability of hitting the target by $A,B$ and $C$ are $\displaystyle \frac{2}{3}\;,\frac{1}{2}$ and $\displaystyle \frac{1}{3}$ respectively. if $A$ is hit, then find the probability that $B$ hits the target and $C$ does not

Assume $P(A)$ probability that $A$ hits the target $\displaystyle \frac{2}{3}$

$P(B)$ probability that $B$ hits the target $\displaystyle \frac{1}{2}$

$P(C)$ probability that $C$ hits the target $\displaystyle \frac{1}{3}$

could some help me

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  • $\begingroup$ Probability is 1. "A" got hit, so if it is not "C" who hit him, it must be B $\endgroup$ Jan 31, 2017 at 8:02
  • $\begingroup$ @Markoff You are ignoring the possibility of C hitting A $\endgroup$ Jan 31, 2017 at 8:23

1 Answer 1

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Let $E_1$, $E_2$ and $E_3$ denote the events: $A $ hits $B $, $B $ hits $A $ and $C $ hits $A $ respectively and $E $ denote the event: $A $ is hit.

As $E_2$ and $E_3$ are independent, we have $$P (E) = P (E_2 \cup E_3) = 1- P (E_2' \cap E_3') = 1-(\frac{1}{2})(\frac {2}{3}) = \frac {2}{3}$$

We need to find $$P ((E_2 \cap E_3') \mid E) = \frac {P (E_2 \cap E_3' \cap E)}{P (E)} $$ We have $E_2 \cap E_3' \cap E = (E_2 \cap E_3') \cap (E_2 \cup E_3) = [(E_2 \cap E_3') \cap E_2] \cup [(E_2 \cap E_3') \cap E_3] = E_2 \cap E_3'$.

The answer thus comes put be $\boxed {\frac {1}{2}} $ as the events $E_2$ and $E_3'$ are independent. Hope it helps.

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