Question

A class has 6 students and the teacher wants to group them randomly such that number of students in each group must be even number only. The probability that 2 particular students A and B are in the same group is equal to

• A. $\frac{1}{3}$
• B. $\frac{17}{31}$
• C. $\frac{12}{31}$
• D. $\frac{11}{31}$

Answer 1

Answer: (D)

$\frac{11}{31}$

Answer 2

The problem asks for the probability that two particular students, A and B, are in the same group, given that 6 students are randomly grouped such that the number of students in each group must be an even number.

First, we determine all possible ways to partition 6 students into groups with only even numbers of students. The possible group sizes are:

1. One group of 6 students.
2. One group of 4 students and one group of 2 students.
3. Three groups of 2 students each.

Next, we calculate the total number of distinct ways to form these partitions:

1. Case 1: One group of 6 students. Number of ways = $\binom{6}{6} = 1$.

2. Case 2: One group of 4 students and one group of 2 students. Number of ways = $\binom{6}{4} \times \binom{2}{2} = 15$.

3. Case 3: Three groups of 2 students each. Since the group sizes are the same, we divide by $3!$ to account for indistinguishable groups. Number of ways = $\frac{\binom{6}{2} \times \binom{4}{2} \times \binom{2}{2}}{3!} = \frac{15 \times 6 \times 1}{6} = 15$.

The total number of distinct valid partitions is $1 + 15 + 15 = 31$.

Now, we find the number of partitions where students A and B are in the same group:

1. Case 1: A and B are in the group of 6. This is the single partition where all 6 students form one group. Number of favorable ways = 1.

2. Case 2: A and B are in the same group of size 4 (from the (4, 2) partition). We need to choose 2 more students from the remaining 4 students (excluding A and B) to join them in the group of 4. Number of ways = $\binom{4}{2} = 6$.

3. Case 3: A and B are in the group of size 2 (from the (4, 2) partition). If A and B form the group of 2, the remaining 4 students must form the group of 4. Number of ways = $\binom{4}{4} = 1$.

4. Case 4: A and B are in the same group of size 2 (from the (2, 2, 2) partition). If A and B form one group of 2, we need to partition the remaining 4 students into two groups of 2. Number of ways = $\frac{\binom{4}{2} \times \binom{2}{2}}{2!} = \frac{6 \times 1}{2} = 3$.

The total number of favorable ways is $1 + 6 + 1 + 3 = 11$.

The probability is the ratio of favorable ways to the total possible ways: Probability = $\frac{11}{31}$.