Three persons A,B and C shoot to hit a target.If in trials,A hits the target 4 times in 5 shots,B hits 3 times in 4 shots and C hits 2 times in 3 trials.Find the probability that :

(a) Exactly two persons hit the target. (b) At least two persons hit the target. This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

$P(A) = \frac{4}{5} P(A') = \frac{1}{5}$  ( Probability of A hitting and not hitting )
$P(B) = \frac{3}{4} P(B') = \frac{1}{4}$
$P(C) = \frac{2}{3} P(C') = \frac{1}{3}$

a) Exactly two persons hit the target

= P( A & B hits & C misses ) + P( A & C hits & B misses ) + P( B & C hits & A misses )

$= P(A)*P(B)*P(C') + P(A)*P(C)*P(B') + P(B)*P(C)*P(A')$

$= \frac{4*3*1}{5*4*3} + \frac{4*2*1}{5*3*4} + \frac{3*2*1}{4*3*5}$

$= \frac{12+8+6}{60} = \frac{13}{30}$

b ) Atleast two persons hit the target

= P(Exactly two hit) + P(all three hit)

For a) we know = P(Exactly two hit) = $\frac{13}{30}$
P(all 3 hit ) = P(A)*P(B)*P(C) = $\frac{4*3*2}{5*4*3} = \frac{12}{30}$

So P(atleast 2 persons hit the target) = $\frac{13}{30} + \frac{12}{30} = \frac{25}{30} = \frac{5}{6}$