 QUESTION

# A can hit a target 3 times in 6 shots, B can hit a target 2 times in 6 shots and C can hit 4 times in 4 shots. They fix a volley. What is the probability that at least 2 shots hit?

Hint: Note that A, B, and C hitting the shots are independent events. Find the possibilities for hitting at least two shots and calculate the probability for independent events using $P(A \cap B \cap C) = P(A)P(B)P(C)$.

It is given that A can hit a target 3 times in 6 shots, hence, the probability of A is given as follows:

$P(A) = \dfrac{3}{6}$

$P(A) = \dfrac{1}{2}............(1)$

Let the event $\bar A$ be the event that A does not hit the target. Then, we have:

$P(\bar A) = 1 - \dfrac{1}{2}$

$P(\bar A) = \dfrac{1}{2}.............(2)$

It is given that B can hit a target 2 times in 6 shots, hence, the probability of B is given as follows:

$P(B) = \dfrac{2}{6}$

$P(B) = \dfrac{1}{3}............(3)$

Let the event $\bar B$ be the event that B does not hit the target. Then, we have:
$P(\bar B) = 1 - \dfrac{1}{3}$

$P(\bar B) = \dfrac{2}{3}............(4)$

It is given that C can hit a target 4 times in 4 shots, hence, the probability of C is given as follows:

$P(C) = \dfrac{4}{4}$

$P(C) = 1............(5)$

Let the event $\bar C$ be the event that C does not hit the target. Then, we have:
$P(\bar C) = 1 - 1$

$P(\bar C) = 0............(6)$

In a volley, we need to find the probability that at least two shots hit the target. We have the following possibilities: Both A and B hit, C does not hit, both A and C hit, B does not hit, both B and C hit, A does not hit, all three hit the target. Hence, the required probability is given as follows:

$P = P(A \cap B \cap \bar C) + P(A \cap \bar B \cap C) + P(\bar A \cap B \cap C) + P(A \cap B \cap C)$

We know that A, B, and C are independent events, hence, the probability of the intersection of events can be written as the product of probabilities of individual events.

$P(A \cap B \cap C) = P(A)P(B)P(C)$

Hence, we have the following.

$P = P(A)P(B)P(\bar C) + P(A)P(\bar B)P(C) + P(\bar A)P(B)P(C) + P(A)P(B)P(C)$

Substituting equation (1) to equation (6), we have:

$P = \dfrac{1}{2}.\dfrac{1}{3}.0 + \dfrac{1}{2}.\dfrac{2}{3}.1 + \dfrac{1}{2}.\dfrac{1}{3}.1 + \dfrac{1}{2}.\dfrac{1}{3}.1$

Simplifying, we have:

$P = \dfrac{1}{6} + \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{6}$

Taking common denominator, we have:

$P = \dfrac{{1 + 2 + 1 + 1}}{6}$

Simplifying, we have:

$P = \dfrac{4}{6}$

$P = \dfrac{2}{3}$

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

Note: You can also find the probability that less than 2 shots hit the target and subtract the value from 1 to get the probability that at least 2 shots hit the target.