Question

From a set comprising of 10 students, four girls 𝐺_𝑖 , 𝑖 = 1, … , 4, and six boys 𝐵_𝑗 , 𝑗 = 1, … , 6, a team of five students is to be formed. The probability that a randomly selected team comprises of 2 girls and 3 boys, with at least one of them to be 𝐵₁ or 𝐵₂ , is equal to

• A. \(\frac{3}{7}\)
• B. \(\frac{6}{7}\)
• C. \(\frac{8}{21}\)
• D. \(\frac{5}{21}\)

Answer 1

The Correct Option is C

To solve this problem, we need to determine the probability that a randomly selected team of 5 students from a group of 10 students (4 girls and 6 boys) comprises 2 girls and 3 boys, with at least one of the boys being either \( B_1 \) or \( B_2 \).

Let's go through the solution step-by-step:

1. Total number of students = 10
   • Girls (\( G_i \), \( i = 1, 2, 3, 4 \)) = 4 
   • Boys (\( B_j \), \( j = 1, 2, ..., 6 \)) = 6
2. Total number of ways to select any 5 students from these 10 students is given by the combination: \(C(10, 5) = \frac{10!}{5! \times 5!} = 252\)
3. We need to form a team of 2 girls and 3 boys.
4. Number of ways to select 2 girls from 4 is: \(C(4, 2) = \frac{4!}{2! \times 2!} = 6\)
5. Number of ways to select 3 boys from 6 is: \(C(6, 3) = \frac{6!}{3! \times 3!} = 20\)
6. Total ways to select 2 girls and 3 boys: \(6 \times 20 = 120\)
7. Now, we need at least one of those boys to be \( B_1 \) or \( B_2 \).
8. First, calculate the number of ways where none of \( B_1 \) or \( B_2 \) are selected:
9. If \( B_1 \) and \( B_2 \) are not selected, we select all 3 boys from the other 4 boys (\( B_3, B_4, B_5, B_6 \)): \(C(4, 3) = \frac{4!}{3! \times 1!} = 4\)
10. So, the number of ways to select 2 girls and 3 boys without \( B_1 \) and \( B_2 \) is: \(6 \times 4 = 24\)
11. Therefore, the number of ways to select 2 girls and 3 boys with at least one of \( B_1 \) or \( B_2 \) is: \(120 - 24 = 96\)
12. Finally, the probability that the selected team includes at least one of \( B_1 \) or \( B_2 \) is: \(\frac{96}{252} = \frac{8}{21}\)

Thus, the probability that a randomly selected team comprises of 2 girls and 3 boys, with at least one being \( B_1 \) or \( B_2 \), is \(\frac{8}{21}\).