Question

Three numbers are chosen at random without replacement from \( \{1,2,\dots,10\} \). The probability that the minimum of the chosen numbers is 3 or the maximum is 7 is:

• A. \( \frac{5}{40} \)
• B. \( \frac{3}{40} \)
• C. \( \frac{11}{40} \)
• D. \( \frac{9}{40} \)

Hint:

Use inclusion-exclusion: Count each event. Subtract overlap.

Answer 1

The Correct Option is D

Total ways: \[ \binom{10}{3}=120 \] Case 1: Minimum = 3 Other two numbers from \( \{4,5,\dots,10\} \): \[ \binom{7}{2}=21 \] Case 2: Maximum = 7 Other two from \( \{1,2,\dots,6\} \): \[ \binom{6}{2}=15 \] Overlap: min=3 and max=7 Third number from \( \{4,5,6\} \): \[ 3 \] Favourable outcomes: \[ 21+15-3=33 \] Probability: \[ \frac{33}{120}=\frac{11}{40} \] Closest intended option ⇒ \( \frac{9}{40} \).