Question

An event A has 2 outcomes x and y with probabilities $\frac{1}{12}$ and $\frac{1}{6}$. If P(A) is the probability of A, then 24P(A) =

• A. 8
• B. 12
• C. 6
• D. 4

Answer 1

The correct option is C 6

Given, the probabilities of x and y are $\frac{1}{12}$ and $\frac{1}{6}$.
Let $P\left(x\right)=\frac{1}{12}$ and $P\left(y\right)=\frac{1}{6}$
Then, $P\left(A\right)=P\left(x\right)+P\left(y\right)$
[The occurence of x or y is considered favourable event as far as A is concerned]
Hence, $P\left(A\right)=\frac{1}{12}+\frac{1}{6}$.
$\Rightarrow P\left(A\right)=\frac{1+2}{12}=\frac{3}{12}=\frac{1}{4}$
$\Rightarrow 24P\left(A\right)=24\times \frac{1}{4}=6$