Question

Assertion :If A, B, C are three events such that $P\left(A\right)=\frac{1}{4}$, $P\left(B\right)=\frac{1}{6}$ & $P\left(C\right)=\frac{2}{3}$ then events A, B, C are mutually exclusive. Reason: If $P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)$ then A, B, C are mutually exclusive events.

• A. Both Assertion & Reason are individually true & Reason is correct explanation of Assertion
• B. Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion
• C. Assertion is true but Reason is false
• D. Assertion is false but Reason is true

Answer 1

The correct option is D Assertion is false but Reason is true

$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)$
$=\frac{1}{4}+\frac{1}{6}+\frac{2}{3}=\frac{13}{12}$ $>1$ which can not be possible
$\Rightarrow$ Assertion (A) is false but Reason (R) is true.