Question

The probability of A $=$ Probability of B $=$ Probability of C $=\frac{1}{4}$
$P\left(A\right)\cap P\left(B\right)\cap P\left(C\right)=0$. $P(B\cap C)=0$ and $P(A\cap C)=\frac{1}{8}$. $P(A\cap B)=0$ the probability that atleast one of the events A, B, C exists is?

• A. $\frac{5}{8}$
• B. $\frac{37}{64}$
• C. $\frac{3}{4}$
• D. $1$

Answer 1

The correct option is A $\frac{5}{8}$

$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cup C)-P(A\cap C)+P(A\cap B\cap C)$

$⟹\frac{3}{4}-\frac{1}{8}$

$\frac{5}{8}$