Question

Let $A,B,C,D$ be independent even such that $P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3},P\left(C\right)=\frac{1}{5}$ and $P\left(D\right)=\frac{1}{6}$. Then the probability that none of $A,B,C$ and $D$ occurs.

• A. $\frac{1}{180}$
• B. $\frac{1}{45}$
• C. $\frac{1}{18}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{180}$

Given,

$P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{3},P\left(C\right)=\frac{1}{5}$ and $P\left(D\right)=\frac{1}{6}$

Since, $A,B,C$ and $D$ are independent events.

$\therefore A^{\prime },B^{\prime },C^{\prime }$ and $D^{\prime }$ are also independent events.

$\therefore P(A^{\prime }\cap B^{\prime }\cap C^{\prime }\cap D^{\prime })=P\left(A^{\prime }\right)\times P\left(B^{\prime }\right)\times P\left(C^{\prime }\right)\times P\left(D^{\prime }\right)=\frac{1}{2}\times \frac{1}{3}\times \frac{1}{5}\times \frac{1}{6}=\frac{1}{180}$

Hence, option (A) is correct.