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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?

Explanation

Solution

Hint: As it is mentioned to find the probability that at least 2 shots hit. We have to use the formula P[(ABC)(ABC)(ABC)(ABC)]P[(A \cap B \cap C) \cup (\overline A \cap B \cap C) \cup (A \cap \overline B \cap C) \cup (A \cap B \cap \overline 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)=36=12 = P(A) = \dfrac{3}{6} = \dfrac{1}{2}.
So, Probability of A not hitting the target P(A)=1P(A)=112=12P(\overline A ) = 1 - P(A) = 1 - \dfrac{1}{2} = \dfrac{1}{2}.
Similarly, Probability of B hitting the target =P(B)=26=13 = P(B) = \dfrac{2}{6} = \dfrac{1}{3}.
So, Probability of B not hitting the target P(B)=1P(B)=113=23P(\overline B ) = 1 - P(B) = 1 - \dfrac{1}{3} = \dfrac{2}{3}.
Probability of C hitting the target =P(C)=44=1 = P(C) = \dfrac{4}{4} = 1.
So, Probability of C not hitting the target P(C)=1P(C)=11=0P(\overline C ) = 1 - P(C) = 1 - 1 = 0.
Required probability=P[(ABC)(ABC)(ABC)(ABC)] = P[(A \cap B \cap C) \cup (\overline A \cap B \cap C) \cup (A \cap \overline B \cap C) \cup (A \cap B \cap \overline C )].
As A, B, C are independent events. So, the required probability will become,

=P(A)P(B)P(C)+P(A)P(B)P(C)+P(A)P(B)P(C)+P(A)P(B)P(C) =12×13×1+12×13×1+12×23×1+12×13×0 =16+16+26+0 =46=23  = P(A)P(B)P(C) + P(\overline A )P(B)P(C) + P(A)P(\overline B )P(C) + P(A)P(B)P(\overline C ) \\\ = \dfrac{1}{2} \times \dfrac{1}{3} \times 1 + \dfrac{1}{2} \times \dfrac{1}{3} \times 1 + \dfrac{1}{2} \times \dfrac{2}{3} \times 1 + \dfrac{1}{2} \times \dfrac{1}{3} \times 0 \\\ = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{2}{6} + 0 \\\ = \dfrac{4}{6} = \dfrac{2}{3} \\\

Hence the required probability is 23\dfrac{2}{3}.

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.