Question

A question paper is split in two groups – A and B groups. A contains 4 questions, each question having an alternative. The group B also contains 4 questions. A student has to answer at least 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

Hint:
 In this question, first, we will check that in how many ways, questions can be attempted by a student in both sections that is section A and section B. After that, we will apply the formula for finding the probability of solving the question by a student. We will apply here, the tactics of solving problems using permutation and combination and probability.

Complete step by step answer:
Let us solve the problem.
We have two groups A and B.
Group A has 4 questions and each question is having an alternative.
Therefore, the number of ways, a student can answer the question in section A:
by answering the question,
by answering the alternatives, and
by skipping the question
Hence, the total numbers of ways are \[{{3}^{4}}=81\]
But, we have a case in which students can skip all the questions.
Thus, the numbers of ways are 81-1=80
In a similar way, in section B, the student has two options that he can answer the question or skip the question.
Therefore, the total numbers of ways to answer in section B are \[{{2}^{4}}-1=15\]
Hence, the required numbers of ways are \[80\times 15=1200\]
Now, we have to find the probability that a student will answer three questions(There are two possibilities if he answered 2 questions from A and 1 question from B and another possibility is that he answered 2 questions from B and 1 question from A.)
\[{}^{4}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{4}{{C}_{1}}\times \left( {}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}+{}^{3}{{C}_{1}} \right)=288\]
Therefore, probability = \[\dfrac{288}{1200}\].

Note:
 One should have proper knowledge in binomial theorem, permutation, and combination, and probability to solve this type of problem. One can make mistake in solving this problem. The most common mistake that is made by a majority of students is double counting.