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Question: Two men hit at a target with probabilities \(\dfrac{1}{2}\) and \(\dfrac{1}{3}\) respectively. What ...

Two men hit at a target with probabilities 12\dfrac{1}{2} and 13\dfrac{1}{3} respectively. What is the probability that exactly one of them hits the target?
A)12A)\dfrac{1}{2}
B)13B)\dfrac{1}{3}
C)16C)\dfrac{1}{6}
D)23D)\dfrac{2}{3}

Explanation

Solution

Probability is the term mathematically with events that occur, which is the number of favorable events that divides the total number of the outcomes.
If we divide the probability and then multiplied with the hundred then we will determine its percentage value.
16\dfrac{1}{6} which means the favorable event is 11 and the total outcome is 66
Formula used:
P=FTP = \dfrac{F}{T}where P is the overall given probability, F is the possible favorable events and T is the total outcomes from the given.
the probability that exactly one of them will hit is P=P(E1)P(E2‾)+P(E2)P(E1‾)P = P({E_1})P(\overline {{E_2}} ) + P({E_2})P(\overline {{E_1}} )

Complete step by step answer:
Let us denote the E1{E_1} to the event when the first man will hit the target, hence we have the given probability is 12\dfrac{1}{2} and also the first man will not hit the target probability is 1−12=121 - \dfrac{1}{2} = \dfrac{1}{2}
These two can be represented as P(E1)=12,P(E1‾)=12P({E_1}) = \dfrac{1}{2},P(\overline {{E_1}} ) = \dfrac{1}{2} respectively.
Let us denote the E2{E_2} to the event when the second man will hit the target, hence we have the given probability is 13\dfrac{1}{3} and also the second man will not hit the target probability is 1−13=231 - \dfrac{1}{3} = \dfrac{2}{3}
These two can be represented as P(E2)=13,P(E2‾)=23P({E_2}) = \dfrac{1}{3},P(\overline {{E_2}} ) = \dfrac{2}{3} respectively.
To find the probability that exactly one of them will hit is P=P(E1)P(E2‾)+P(E2)P(E1‾)P = P({E_1})P(\overline {{E_2}} ) + P({E_2})P(\overline {{E_1}} )
Substitute all the values in the formula we get P=P(E1)P(E2‾)+P(E2)P(E1‾)P = P({E_1})P(\overline {{E_2}} ) + P({E_2})P(\overline {{E_1}} )
$$$\Rightarrow \dfrac{1}{2} \times \dfrac{2}{3} + \dfrac{1}{3} \times \dfrac{1}{2}Furthersolvingweget Further solving we getP = \dfrac{1}{2} \times \dfrac{2}{3} + \dfrac{1}{3} \times \dfrac{1}{2} \Rightarrow \dfrac{1}{2}$$ is the probability that exactly one of them will hit the target.

So, the correct answer is “Option A”.

Note:
If we divide the probability and then multiplied with the hundred then we will determine its percentage value.
Hence, we have 12=50%\dfrac{1}{2} = 50\% is the percentage of exactly one of them that will hit the target.
Note that P(E1)+P(E1‾)=1P({E_1}) + P(\overline {{E_1}} ) = 1 is a result. Where P(E1‾)P(\overline {{E_1}} ) not possible events according to the given.