Question

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?

Answer 1

Hint: First find the probabilities of $A,B$ and $C$, that they hit the target and they can’t hit the target and then use them to find the probability that at least 2 shots hit the target.

Complete step-by-step answer:
It is give that “A” can hit a target 3 times in 6 shots, then the probability is given as:
$P\left( A \right)$ $= \dfrac{3}{6} = \dfrac{1}{2}$
We know that the sum of probabilities of the possible events is always 1, therefore we can find the probability that A can not hit the target by subtracting the probability of hitting the target by 1.
Then the probability that “A” can not hit the target is given as:
$P\left( {\overline A } \right)$ $= 1 - \dfrac{1}{2} = \dfrac{1}{2}$
Now, it is given that “B” can hit the target 2 times in 6 shots then the probability is given as:
$P\left( B \right)$ $= \dfrac{2}{6} = \dfrac{1}{3}$
Then the probability that “B” can not hit the target is given as:
$P\left( {\overline B } \right)$ $= 1 - \dfrac{1}{3} = \dfrac{2}{3}$
Now, it is given that “C” can hit the target 4 times in 4 shots then the probability is given as:
$P\left( C \right)$ $= \dfrac{4}{4} = 1$
Then the probability that C cannot hit the target is given as:
$P\left( {\overline C } \right) = 1 - 1 = 0$
It can be seen that A, B, and C are independent events therefore; the probability that at least 2 shots hit the target has the four possibilities.
(1) A and B hit the target and C does not hit the target;
(2) B and C hit the target but A does not hit the target;
(3) A and C hit the target but B does not hit the target;
(4) All A,B and C hit the target.
Therefore, the probability that at least 2 shots hit the target is the sum of all possible probabilities.
Assume that $P\left( R \right)$ is the probability that at least 2 shots hit the target, then the probability is given as:
$P\left( R \right) = {\text{Probability that A and B hit the target not C}} + {\text{Probability that A and C hit the target not B}} \\\ \+ {\text{Probability that B and C hit the target not A }} + {\text{ A, B and C hit the target}} \\\$
$P\left( R \right) = P\left( A \right)P\left( B \right)P\left( {\overline C } \right) + P\left( A \right)P\left( {\overline B } \right)P\left( C \right) + P\left( {\overline A } \right)P\left( B \right)P\left( C \right) + P\left( A \right)P\left( B \right)P\left( C \right)$
Substitute the values of the probabilities in the above expression:
P\left( R \right) = \dfrac{1}{2} \times \dfrac{1}{3} \times 0 + \dfrac{1}{2} \times \dfrac{2}{3} \times 1 + \dfrac{1}{2} \times \dfrac{1}{3} \times 1 + \dfrac{1}{2} \times \dfrac{1}{3} \times 1\] P\left( R \right) = 0 + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6}$$P\left( R \right) = \dfrac{2}{3}$

Therefore, the probability that at least 2 shots hit the target is$\dfrac{2}{3}$.

Note: It is given that at least 2 shots hit the target then the possible event also contains the event in which more than 2 shots hit the target, and there is one possible event in which all $A,B$ and $C$ hits the target together.