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 .