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?
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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