Question

A question paper is split in two groups $-A$ and $B$. The group $A$ contains $4$ questions, each question having an alternative. The group $B$ also contains $4$ questions. A student has to answer atleast one question from each group and he can answer up to $8$ questions. What is the probability that a student will answer $3$ questions?

Answer 1

Number of the ways we can answer the question in section $A$

1. we can answer the ques

2. we can answer the alternative

3. we can skip

Therefore, the total number of ways $3^4=81$.

But there is one case where we have skipped all the questions.

Thus number of ways $=81-1=80$.

Similarly, in section $B$ we have two options either we can answer the ques or we can skip the ques.

$\therefore$ total number of ways to answer $=2^4-1=15$

Hence, required number of ways $80\times 15=1200$

If student has answered $3$ questions, then there are $2$ possibilities either he answered $2$ questions from $A$ or from $B$.

${}^4{C}_1{\cdot }^2{C}_1{\cdot }^4{C}_1{(}^3{C}_1{\cdot }^2{C}_1{+}^3{C}_1)=288$

Therefore, probability$=\frac{288}{1200}$