Understanding the Probability Problem

The question asks for the probability of selecting 3 students from a council such that exactly two students are from the same school, and the third student is from a different school. The council has 10 members distributed across three schools: 2 from Engineering, 3 from Science, and 5 from Arts.

Calculating Total Possible Selections

First, determine the total number of ways to select any 3 students from the 10 council members. This is a combination problem, calculated as follows:

Total ways = $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$.

Calculating Favorable Selections

Next, identify the scenarios where exactly two students are from the same school and one is from a different school. There are three possible cases based on which school the pair comes from:

• Case 1: Two Engineering students and one student from a different school.
  • Ways to choose 2 Engineering students: $\binom{2}{2} = 1$.
  • Ways to choose 1 student from the remaining 8 (3 Science + 5 Arts): $\binom{8}{1} = 8$.
  • Total ways for Case 1: $1 \times 8 = 8$.
• Case 2: Two Science students and one student from a different school.
  • Ways to choose 2 Science students: $\binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3$.
  • Ways to choose 1 student from the remaining 7 (2 Engineering + 5 Arts): $\binom{7}{1} = 7$.
  • Total ways for Case 2: $3 \times 7 = 21$.
• Case 3: Two Arts students and one student from a different school.
  • Ways to choose 2 Arts students: $\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$.
  • Ways to choose 1 student from the remaining 5 (2 Engineering + 3 Science): $\binom{5}{1} = 5$.
  • Total ways for Case 3: $10 \times 5 = 50$.

The total number of favorable selections is the sum of the ways from these three cases:

Total Favorable Ways = $8 + 21 + 50 = 79$.

Determining the Probability

The probability is the ratio of the total favorable selections to the total possible selections:

Probability = $\frac{\text{Total Favorable Ways}}{\text{Total Possible Ways}} = \frac{79}{120}$.