Question

A can hit a target 3 times in 6 shots, B can hit 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 are hit?

Answer 1

Hint : We can calculate respective probabilities for both hitting and non-hitting a target by A, B and C and then substitute these values in the required probability. The required probability will be based on hitting 2 or more shots by anyone.
$P = \dfrac{f}{T}$ where,
P = Probability
f = Favorable outcomes
T = Total outcomes
Sum of probabilities of happening and non-happening of an event is 1
$P(A) + P(\overline A ) = 1 \\\ P(\overline A ) = 1 - P(A) \\\$

Complete step-by-step answer :
Favorable outcomes (f) = 3 [hits the target]
Total outcomes (T) = 6 [number of shots]
Probability that A hits the target
$\Rightarrow P(A) = \dfrac{3}{6}$
$\Rightarrow P(A) = \dfrac{1}{2}$
Probability that A does not hit the target
$\Rightarrow P(\overline A ) = 1 - \dfrac{1}{2}$
$\Rightarrow P(\overline A ) = \dfrac{1}{2}$
Target hit by B:
Favorable outcomes (f) = 2 [hits the target]
Total outcomes (T) = 6 [number of shots]
Probability that B hits the target
$\Rightarrow P(B) = \dfrac{2}{6}$
$\Rightarrow P(B) = \dfrac{1}{3}$
Probability that B does not hit the target
$\Rightarrow P(\overline B ) = 1 - \dfrac{1}{3}$
$\Rightarrow P(\overline B ) = \dfrac{2}{3}$
Target hit by C:
Favorable outcomes (f) = 4 [hits the target]
Total outcomes (T) = 4 [number of shots]
Probability that C hits the target
$\Rightarrow P(C) = \dfrac{4}{4}$
P (C) = 1
Probability that C does not hit the target
$\Rightarrow P(\overline C ) = 1 - 1$
$\Rightarrow P(\overline C ) = 0$
Now, the required probability of hitting at least two shots will be on the basis:
A and B hits but C does not or
A and C hits but B does not or
B and C hits but A does not or
A and B hits but C does not or
A, B and C all hit the shots (as more than two are possible)
$\Rightarrow P = P(A).P(B).P(\overline C ) + P(A).P(C).P(\overline B ) + P(B).P(C).P(\overline A ) + P(A).P(B).P(C)$
Substituting the values, we get:
$\Rightarrow P = \dfrac{1}{2} \times \dfrac{1}{3} \times 0 + \dfrac{1}{2} \times 1 \times \dfrac{2}{3} + \dfrac{1}{3} \times 1 \times \dfrac{1}{2} + \dfrac{1}{2} \times \dfrac{1}{3} \times 1 \\\ \Rightarrow P = \dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{6} \\\ \Rightarrow P = \dfrac{2}{3} \\\$

Note : In probabilities, if two events have ‘or’ amidst them, then in the mathematical calculation it will be ‘+’ and in case of ‘and’ it will be ‘X’
If the probability of any event is 1, it means that that event will surely occur whereas an event with probability 0 is impossible to occur.
We use a bar over the probability symbol to show negation (not happening of an event)