A student council of 10 members consists of two students from engineering School, 3 students from science School and 5 students from art School. the uni. admin select 3 students from the council at random. What is the chance that out of the selected students, two belongs to the same school & the 3rd belongs to a different school:

The correct answer is 79/120.

To find the probability that out of the three selected students, exactly two belong to the same school and the third belongs to a different school, we can analyze the composition of the student council and the possible selection outcomes.

Step 1: Identify the council composition and total ways of selection
The student council consists of 10 members in total:
- Engineering School (E): 2 students
- Science School (S): 3 students
- Art School (A): 5 students

We are selecting 3 students at random from these 10 members. The total number of ways to choose 3 students from 10 is given by the combination formula:

$$\text{Total ways}={}_{10}C_3=\frac{10\times 9\times 8}{3\times 2\times 1}=120$$

Step 2: Use the complementary counting method
When selecting 3 students from the council, there are only three possible categories of outcomes:
1. All 3 students belong to the same school.
2. All 3 students belong to three different schools (one from each school).
3. Exactly 2 students belong to the same school and the 3rd belongs to a different school (our desired event).

Therefore, we can find the number of favorable ways for our desired event by subtracting the outcomes of Category 1 and Category 2 from the total number of ways (120).

Step 3: Calculate Category 1 (All 3 students from the same school)
Since there are only 2 students from the Engineering school, we cannot choose 3 Engineering students. We can only choose 3 students from the Science school or 3 students from the Art school:
- Ways to choose 3 Science students: ³C₃ = 1 way
- Ways to choose 3 Art students: ⁵C₃ = 10 ways

$$\text{Total ways for Category 1}=1+10=11$$

Step 4: Calculate Category 2 (All 3 students from different schools)
To get 3 students from different schools, we must select exactly 1 student from Engineering, 1 from Science, and 1 from Art:

$$\text{Total ways for Category 2}={}_{2}C_1\times {}_{3}C_1\times {}_{5}C_1=2\times 3\times 5=30$$

Step 5: Calculate the number of favorable ways for Category 3
Subtracting the ways of Category 1 and Category 2 from the total selection ways:

$$\text{Favorable ways}=120-11-30=79$$

Step 6: Compute the final probability
The probability of selecting exactly two students from the same school and the third from a different school is:

$$\text{Probability}=\frac{\text{Favorable ways}}{\text{Total ways}}=\frac{79}{120}$$