Question
Question: A can hit a target 3 times in 6 shots, B: 2 times in 6 shots, C: 4 times in 4 shots. They fix a voll...
A can hit a target 3 times in 6 shots, B: 2 times in 6 shots, C: 4 times in 4 shots. They fix a volley. What is the probability that at least 2 shots hit?
Solution
Hint: As it is mentioned to find the probability that at least 2 shots hit. We have to use the formula P[(A∩B∩C)∪(A∩B∩C)∪(A∩B∩C)∪(A∩B∩C)] for required probability. Observe that A, B and C are independent events.
Complete step-by-step answer:
We know that probability is nothing but a ratio of number of favourable cases to number of total cases. Using this information, for A, B and C.
Probability of A hitting the target =P(A)=63=21.
So, Probability of A not hitting the target P(A)=1−P(A)=1−21=21.
Similarly, Probability of B hitting the target =P(B)=62=31.
So, Probability of B not hitting the target P(B)=1−P(B)=1−31=32.
Probability of C hitting the target =P(C)=44=1.
So, Probability of C not hitting the target P(C)=1−P(C)=1−1=0.
Required probability=P[(A∩B∩C)∪(A∩B∩C)∪(A∩B∩C)∪(A∩B∩C)].
As A, B, C are independent events. So, the required probability will become,
Hence the required probability is 32.
Note: When we say A, B, C are independent events, it means, when even A is performed, it has nothing to do with B, C. In the same manner when B is performed it has nothing to do with other twos and similar with C.