Question

The probability of solving a question by three students are $\frac{1}{2},\frac{1}{4}$ and $\frac{1}{6}$ respectively. probability of question being solved will be :

• A. $\frac{33}{48}$
• B. $\frac{35}{48}$
• C. $\frac{31}{48}$
• D. $\frac{37}{48}$

Answer 1

The correct option is A $\frac{33}{48}$

Let,

$P\left(A\right)=\frac{1}{2},P\left(B\right)=\frac{1}{4}$ and $P\left(C\right)=\frac{1}{6}$

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

$\therefore$Probability of solving the question is,

$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\times \frac{1}{4}-\frac{1}{4}\times \frac{1}{6}-\frac{1}{2}\times \frac{1}{6}+\frac{1}{2}\times \frac{1}{4}\times \frac{1}{6}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}-\frac{1}{8}-\frac{1}{24}-\frac{1}{12}+\frac{1}{48}=\frac{33}{48}$

Hence, option (A) is correct.