Question:
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?
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?
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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
- \( \dfrac{79}{120} \)
- \( \dfrac{2}{120} \)
- \( \dfrac{89}{120} \)
- \( \dfrac{69}{120} \)
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The Correct Option is A
Solution and Explanation
Step 1: Find the total number of ways to choose 3 students.
Out of \( 10 \) members, choosing any \( 3 \) can be done in \( \binom{10}{3} = 120 \) ways, and every selection is equally likely.
Step 2: Count selections with two from Engineering and one from elsewhere.
Engineering has only \( 2 \) students, so both must be picked: \( \binom{2}{2}=1 \) way. The third student comes from the remaining \( 8 \) (science and arts combined): \( \binom{8}{1}=8 \) ways.
This case gives \( 1 \times 8 = 8 \) favourable selections.
Step 3: Count selections with two from Science and one from elsewhere.
Choose \( 2 \) of the \( 3 \) science students: \( \binom{3}{2}=3 \) ways. The third student comes from the other \( 7 \) (engineering and arts): \( \binom{7}{1}=7 \) ways.
This case gives \( 3 \times 7 = 21 \) favourable selections.
Step 4: Count selections with two from Arts and one from elsewhere.
Choose \( 2 \) of the \( 5 \) arts students: \( \binom{5}{2}=10 \) ways. The third student comes from the other \( 5 \) (engineering and science): \( \binom{5}{1}=5 \) ways.
This case gives \( 10 \times 5 = 50 \) favourable selections.
Step 5: Add the three cases and form the probability.
Total favourable ways \( = 8+21+50 = 79 \).
Probability \( = 79/120 \).
Final Answer:
This is the chance that two selected students share a school and the third comes from a different one. \[ \boxed{P = \dfrac{79}{120}} \]
Out of \( 10 \) members, choosing any \( 3 \) can be done in \( \binom{10}{3} = 120 \) ways, and every selection is equally likely.
Step 2: Count selections with two from Engineering and one from elsewhere.
Engineering has only \( 2 \) students, so both must be picked: \( \binom{2}{2}=1 \) way. The third student comes from the remaining \( 8 \) (science and arts combined): \( \binom{8}{1}=8 \) ways.
This case gives \( 1 \times 8 = 8 \) favourable selections.
Step 3: Count selections with two from Science and one from elsewhere.
Choose \( 2 \) of the \( 3 \) science students: \( \binom{3}{2}=3 \) ways. The third student comes from the other \( 7 \) (engineering and arts): \( \binom{7}{1}=7 \) ways.
This case gives \( 3 \times 7 = 21 \) favourable selections.
Step 4: Count selections with two from Arts and one from elsewhere.
Choose \( 2 \) of the \( 5 \) arts students: \( \binom{5}{2}=10 \) ways. The third student comes from the other \( 5 \) (engineering and science): \( \binom{5}{1}=5 \) ways.
This case gives \( 10 \times 5 = 50 \) favourable selections.
Step 5: Add the three cases and form the probability.
Total favourable ways \( = 8+21+50 = 79 \).
Probability \( = 79/120 \).
Final Answer:
This is the chance that two selected students share a school and the third comes from a different one. \[ \boxed{P = \dfrac{79}{120}} \]
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