Question

Let A, B, C be 3 independent events such that $P\left( A \right) = \dfrac{1}{3}$ , $P\left( B \right) = \dfrac{1}{2}$ , $P\left( C \right) = \dfrac{1}{4}$ . Then probability of exactly 2 events occurring out of 3 events is
A $\dfrac{1}{4}$
B $\dfrac{9}{{24}}$
C $\dfrac{3}{4}$
D none of these

Answer 1

Hint : A probability event can be defined as a set of outcomes of an experiment. The probability of occurrence of any event lies between 0 and 1. The occurrences of an event is determined by the number of choices or outcomes for two independent events and multiplying them together. The product of these outcomes will give you the total number of outcomes for each event.

Complete step-by-step answer :
Let us write the given data:
A, B, C are the three independent events, such that:
$P\left( A \right) = \dfrac{1}{3}$ ,
$P\left( B \right) = \dfrac{1}{2}$ ,
$P\left( C \right) = \dfrac{1}{4}$
Hence, we need to find the probability of exactly 2 events occurring out of 3 events i.e.,
$P = P\left( A \right)P\left( B \right)P\left( {\overline C } \right) + P\left( A \right)P\left( {\overline B } \right)P\left( C \right) + P\left( {\overline A } \right)P\left( B \right)P\left( C \right)$
The value of $P\left( {\overline A } \right),P\left( {\overline B } \right),P\left( {\overline C } \right)$ is:
$P\left( {\overline A } \right) = 1 - P\left( A \right)$
Substitute the given value of $P\left( A \right)$ i.e.,
$\Rightarrow$ $P\left( {\overline A } \right)$ = $1 - \dfrac{1}{3}$ = $\dfrac{{3 - 1}}{3}$ = $\dfrac{2}{3}$
$P\left( {\overline B } \right) = 1 - P\left( B \right)$
Substitute the given value of $P\left( B \right)$ i.e.,
$\Rightarrow$ $P\left( {\overline B } \right) = 1 - \dfrac{1}{2} = \dfrac{{2 - 1}}{2} = \dfrac{1}{2}$
$P\left( {\overline C } \right) = 1 - P\left( C \right)$
Substitute the given value of $P\left( C \right)$ i.e.,
$\Rightarrow$ $P\left( {\overline C } \right) = 1 - \dfrac{1}{4} = \dfrac{{4 - 1}}{4} = \dfrac{3}{4}$
Now, substitute the values of P(A), P(B) and P(C) as:
$P = P\left( A \right)P\left( B \right)P\left( {\overline C } \right) + P\left( A \right)P\left( {\overline B } \right)P\left( C \right) + P\left( {\overline A } \right)P\left( B \right)P\left( C \right)$
P = $\left( {\dfrac{1}{3} \times \dfrac{1}{2} \times \dfrac{3}{4}} \right) + \left( {\dfrac{1}{3} \times \dfrac{1}{2} \times \dfrac{1}{4}} \right) + \left( {\dfrac{2}{3} \times \dfrac{1}{2} \times \dfrac{1}{4}} \right)$
P = $\dfrac{1}{4}$
Hence, option A is the right answer.
So, the correct answer is “Option A”.

Note : The probability of an event cannot be greater than 1 since the number of trials in which the event cannot be greater than the total number of trials. The number of favourable outcomes to the total number of outcomes is defined as the probability of occurrence of any event. So, the probability that an event will occur is given as:
P(E) = Number of Favourable Outcomes with respect to the Total Number of Outcomes of the event in the probability.