Question:hard

A student council of \( 10 \) members has two students from the engineering school, three students from the science school, and five students from the arts school. The university administration picks three students from the council at random. What is the chance that among the three chosen students, two belong to the same school and the third belongs to a different school?

Show Hint

Split the outcomes into all same school, all different schools, and two same plus one different, then count each case.
Updated On: Jul 27, 2026
  • \( \dfrac{79}{120} \)
  • \( \dfrac{2}{120} \)
  • \( \dfrac{89}{120} \)
  • \( \dfrac{69}{120} \)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: List the only three ways to pick 3 students by school pattern.
With group sizes 2, 3 and 5, any group of 3 students chosen from the council must fall into one of three patterns: all three from one school, one from each of the three schools, or two from one school plus one from another. These three patterns cover every possible case, so their probabilities must add to 1.

Step 2: Count the all three from one school case.
Engineering only has 2 members, so it cannot supply 3. Science gives $\binom{3}{3}=1$ way, Arts gives $\binom{5}{3}=10$ ways. Total: $1+10=11$ ways.

Step 3: Count the one from each school case.
Pick 1 from Engineering, 1 from Science, 1 from Arts: $2 \times 3 \times 5 = 30$ ways.

Step 4: Subtract both from the total to get the target case.
Total ways are $\binom{10}{3}=120$. The remaining pattern, two from one school and the third from a different school, must be $120 - 11 - 30 = 79$ ways.

Step 5: Form the probability.
$P = \dfrac{79}{120}$, the same value the direct case by case count gives.

Final Answer:
Subtracting the all same and all different patterns from the total confirms the answer. \[ \boxed{P = \dfrac{79}{120}} \]
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