# A can hit a target 3 times in 6 shots, B:2 times in 6 shots and C:4 times in 4 shots. They fix a volley. What is the probability that at least 2 shots hit?

Given:

A hits a target 3 times out of 6 shots

B hits a target 2 times out of 6 shots

C hits a target 4 times out of 4 shots

P(TA) = P(A hits target)

P(TB) = P(B hits target)

P(TC) = P(C hits target)

P(NA) = P(A doesn’t hits target)

P(NB) = P(B doesn’t hits target)

P(NC) = P(C doesn’t hits target)

.

It is told that the target is to be hit by at least two shots. This is only possible when at least two of the Persons hits the target.

P(M) = P(TATBNC) + P(TANBTC) + P(NATBTc) + P(TATBTC)

Since hitting by a person is independent the probabilities will multiply each other.

P(M) = (P(TA)P(TB)P(NC)) + (P(TA)P(NB)P(TC)) + (P(NA)P(TB)P(Tc)) + (P(TA)P(TB)P(TC))

.

The required probability is .

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