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 their maximum is 7, is:

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

Hint:

Remember the inclusion-exclusion principle for 'or' probabilities.

Answer 1

The Correct Option is C

Step 1: Total combinations of 3 numbers.
\n\( \binom{10}{3} = 120 \)\n\n
Step 2: Combinations with a minimum of 3.
\nSelect 2 numbers from \( \{4, 5, 6, 7, 8, 9, 10\} \): \( \binom{7}{2} = 21 \)\n\n
Step 3: Combinations with a maximum of 7.
\nSelect 2 numbers from \( \{1, 2, 3, 4, 5, 6\} \): \( \binom{6}{2} = 15 \)\n\n
Step 4: Combinations with a minimum of 3 AND a maximum of 7.
\nThe set is \( \{3, x, 7\} \) where \( x \in \{4, 5, 6\} \): \( \binom{3}{1} = 3 \)\n\n
Step 5: Apply inclusion-exclusion principle.
\nFavorable outcomes = \( 21 + 15 - 3 = 33 \)\n\n
Step 6: Calculate the probability.
\n\( \frac{33}{120} = \frac{11}{40} \)