Three persons P, Q and R independently try to hit a target. I the probabilities of their hitting the target are \[{3 \over 4},{1 \over 2}\] and \[{5 \over 8}\] respectively, then the probability that the target is hit by P or Q but not by R is :
Solutions · 1
Official solution
We have the following probabilities:
\[\bullet\] The probability that the target is hit by the person P is \[{3 \over 4}\].
\[\bullet\] The probability that the target is not hit by the person P is \[1 - {3 \over 4} = {1 \over 4}\].
\[\bullet\] The probability that the target is hit by the person Q is \[{1 \over 2}\].
\[\bullet\] The probability that the target is not hit by the person Q is \[1 - {1 \over 2} = {1 \over 2}\].
\[\bullet\] The probability that the target is hit by the person R is \[{5 \over 8}\].
\[\bullet\] The probability that the target is not hit by the person R is \[1 - {5 \over 8} = {3 \over 8}\].
Here, we have used the fact that if the probability of occurrence of an event is p, then the probability of non-occurrence of an event is q=1−p.
Therefore, the probability that the target is hit by P or Q and not by R is
(Probability that the target is hit by P and not by Q and R) + (Probability that the target is hit by Q and not by P and R) + (Probability that the target is hit by both P and Q and not by R)
\[ = \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{1 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right)\]
\[ = {9 \over {64}} + {3 \over {64}} + {9 \over {64}} = {{9 + 3 + 9} \over {64}} = {{21} \over {64}}\]