A, B, C are aiming to shoot a balloon. A will | vedclass

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(C) Let $P(A), P(B), P(C)$ be the probabilities of $A, B, C$ hitting the balloon respectively.$P(A) = \frac{4}{6} = \frac{2}{3}$, so $P(\bar{A}) = 1 -

Vedclass · Accepted Answer

$\frac{32}{45}$

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(C) Let $P(A), P(B), P(C)$ be the probabilities of $A, B, C$ hitting the balloon respectively.$P(A) = \frac{4}{6} = \frac{2}{3}$, so $P(\bar{A}) = 1 - \frac{2}{3} = \frac{1}{3}$.$P(B) = \frac{3}{5}$, so $P(\bar{B}) = 1 - \frac{3}{5} = \frac{2}{5}$.$P(C) = \frac{2}{3}$, so $P(\bar{C}) = 1 - \frac{2}{3} = \frac{1}{3}$.At least two hit the balloon means exactly two hit $OR$ all three hit.Probability (exactly two hit) = $P(A)P(B)P(\bar{C}) + P(A)P(\bar{B})P(C) + P(\bar{A})P(B)P(C)$$= (\frac{2}{3} 	imes \frac{3}{5} 	imes \frac{1}{3}) + (\frac{2}{3} 	imes \frac{2}{5} 	imes \frac{2}{3}) + (\frac{1}{3} 	imes \frac{3}{5} 	imes \frac{2}{3})$$= \frac{6}{45} + \frac{8}{45} + \frac{6}{45} = \frac{20}{45} = \frac{4}{9}$.Probability (all three hit) = $P(A)P(B)P(C) = \frac{2}{3} 	imes \frac{3}{5} 	imes \frac{2}{3} = \frac{12}{45} = \frac{4}{15}$.Total probability = $\frac{20}{45} + \frac{12}{45} = \frac{32}{45}$.

$A, B, C$ are aiming to shoot a balloon. $A$ will succeed $4$ times out of $6$ attempts. The chance of $B$ to shoot the balloon is $3$ out of $5$ and that of $C$ is $2$ out of $3$. If the three aim to shoot the balloon simultaneously, then the probability that at least two of them hit the balloon is

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