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 number of ways to choose 3 numbers.
\( \binom{10}{3} = 120 \)
Step 2: Number of ways minimum is 3.
Choose 2 numbers from \( \{4, 5, 6, 7, 8, 9, 10\} \): \( \binom{7}{2} = 21 \)
Step 3: Number of ways maximum is 7.
Choose 2 numbers from \( \{1, 2, 3, 4, 5, 6\} \): \( \binom{6}{2} = 15 \)
Step 4: Number of ways minimum is 3 AND maximum is 7.
The set is \( \{3, x, 7\} \) where \( x \in \{4, 5, 6\} \): \( \binom{3}{1} = 3 \)
Step 5: Use inclusion-exclusion.
Number of favorable outcomes = \( 21 + 15 - 3 = 33 \)
Step 6: Calculate the probability.
\( \frac{33}{120} = \frac{11}{40} \)