Question

If $\text{A}$ and $\text{B}$ are two events such that$\text{P(AUB)}$$+$$\text{P(A∩B)}$$=$$\frac{7}{8}$, and $\text{P(A)}$$=$$\text{2P(B)}$, then $\text{P(A)}$$=$

• A. $\frac{7}{12}$
• B. $\frac{7}{24}$
• C. $\frac{5}{12}$
• D. $\frac{17}{24}$

Answer 1

The correct option is A

$\frac{7}{12}$

Explantion for the correct option:

Finding the value of $\mathbf{\text{P(A)}}$:

$\text{P(AUB)}$$+$$\text{P(A∩B)}$$=$$\frac{7}{8}$ and

$\text{P(A)}$$=$$\text{2P(B)}$ $....\left(1\right)$

We know that, $\text{P(AUB)}$$=$$\text{P(A)}$$+$$\text{P(B)}$$-$$\text{P(A∩B)}$

$\Rightarrow$ $\text{P(AUB)}$$+$$\text{P(A∩B)}$$=$$\text{P(A)}$$+$$\text{P(B)}$

$=$$\text{P(A)}$$+$$\frac{P\left(A\right)}{2}$ ( from $\left(1\right)$)

$\Rightarrow$ $\frac{7}{8}$ $=$$\frac{3P\left(A\right)}{2}$

$\mathbf{\therefore }$$\mathbf{\text{P(A)}}$ $\mathbf{=}$ $\frac{\mathbf{7}}{\mathbf{12}}$

Hence, option $\mathbf{"}\mathbf{\text{A"}}$ is correct.