Question

In four schools \( B_1, B_2, B_3, B_4 \), the number of students is given as follows: \[ B_1 = 12, \quad B_2 = 20, \quad B_3 = 13, \quad B_4 = 17 \] A student is selected at random from any of the schools. The probability that the student is from school \( B_2 \) is:

• A. \( \frac{6}{31} \)
• B. \( \frac{10}{31} \)
• C. \( \frac{13}{62} \)
• D. \( \frac{17}{62} \)

Hint:

To find the probability of selecting an element from a specific group, use the ratio of favorable outcomes to total outcomes.

Answer 1

The Correct Option is B

Step 1: Calculate the Total Number of Students \[ \text{Total students} = 12 + 20 + 13 + 17 = 62 \]
Step 2: Determine the Probability Since one student is selected at random, the total number of ways to select one student is: \[ \text{Total Outcomes} = \binom{62}{1} = 62 \] The number of students in school \( B_2 \) is: \[ B_2 = 20 \] So the number of ways to select a student from \( B_2 \) is: \[ \binom{20}{1} = 20 \]
Step 3: Compute the Required Probability \[ P(B_2) = \frac{\binom{20}{1}}{\binom{62}{1}} = \frac{20}{62} = \frac{10}{31} \] Final Answer: The probability that the selected student is from \( B_2 \) is: \[ \boxed{\frac{10}{31}} \]