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 volley. What is the probability that at least 2 shots hit?

Answer 1

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

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.