Q. 215.0( 2 Votes )

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?

Answer :

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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