Question

If $A$ and $B$ are two events such that $P\left(A\right)=\frac{1}{4}$ and $P\left(B\right)=P$, the value of $P$ is not ______, if $A\subset B$.

• A. $\frac{1}{12}$
• B. $\frac{3}{4}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is A $\frac{1}{12}$

Let $A$ and $B$ be two events such that $P\left(A\right)=\frac{1}{4},P\left(B\right)=p$ and $A\subset B$.

We have to find which value $p$ does not hold.

We know that, $A\subset B$ if and only if $P\left(A\right)<P\left(B\right)$

Since we have$A<B$, we get

$P\left(A\right)<P\left(B\right)$

Substitute the values of $P\left(A\right)$ and $P\left(B\right)$.

$\Rightarrow \frac{1}{4}<P\left(B\right)$

$\Rightarrow \frac{1}{4}<p$

Consider the values given in the options, $\frac{1}{12},\frac{3}{4},\frac{2}{3},\frac{1}{3}$

Test the above $p$ values with $\frac{1}{4}<p$

$\frac{1}{4}>\frac{1}{12},\frac{1}{4}<\frac{3}{4},\frac{1}{4}<\frac{2}{3},\frac{1}{4}<\frac{1}{3}$.

Thus the only value which does not satisfy $\frac{1}{4}<p$ is $\frac{1}{12}$.

Hence, if $A$ and $B$ are two events such that $P\left(A\right)=\frac{1}{4}$ and $P\left(B\right)=p$, the value of $p$ is not $\frac{1}{12}$, if $A\subset B$.