Question:
Three numbers are chosen at random from \{1, 2, ..., 10\}. The probability that the minimum of the chosen numbers is 3 or their maximum is 7, is
Three numbers are chosen at random from \{1, 2, ..., 10\}. The probability that the minimum of the chosen numbers is 3 or their maximum is 7, is
Show Hint
Don't forget to subtract the intersection event \( P(A \cap B) \). Failing to subtract the overlapping choices where the minimum is 3 and the maximum is 7 will cause you to double-count them, leading to an incorrect inflated answer.
Updated On: Jun 8, 2026
- \( \frac{11}{40} \)
- \( \frac{3}{10} \)
- \( \frac{3}{4} \)
- \( \frac{13}{40} \)
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The Correct Option is A
Solution and Explanation
Concept:
According to the principle of inclusion-exclusion for two events \( A \) and \( B \):
\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]
The total number of ways to select 3 numbers from 10 is \( \binom{10}{3} = \frac{10 \times 9 \times 8}{6} = 120 \).
Step 1: Finding Event A (Minimum number is 3).
If the minimum number must be 3, then 3 is fixed as one of the chosen numbers. The remaining 2 numbers must be chosen from the values strictly greater than 3, which are \( \{4, 5, 6, 7, 8, 9, 10\} \) (7 numbers): \[ \text{Ways for } A = \binom{7}{2} = 21 \]
Step 2: Finding Event B (Maximum number is 7).
If the maximum number must be 7, then 7 is fixed as one of the chosen numbers. The remaining 2 numbers must be chosen from the values strictly less than 7, which are \( \{1, 2, 3, 4, 5, 6\} \) (6 numbers): \[ \text{Ways for } B = \binom{6}{2} = 15 \]
Step 3: Finding Event \( A \cap B \) (Minimum is 3 AND Maximum is 7).
If the minimum is 3 and the maximum is 7, both 3 and 7 are fixed. The remaining 1 number must be chosen from the values strictly between 3 and 7, which are \( \{4, 5, 6\} \) (3 numbers): \[ \text{Ways for } A \cap B = \binom{3}{1} = 3 \]
Step 4: Combining everything using the inclusion-exclusion principle.
\[ \text{Total Favorable Ways} = 21 + 15 - 3 = 33 \] \[ \text{Probability} = \frac{33}{120} = \frac{11}{40} \] This matches option (A) exactly.
Step 1: Finding Event A (Minimum number is 3).
If the minimum number must be 3, then 3 is fixed as one of the chosen numbers. The remaining 2 numbers must be chosen from the values strictly greater than 3, which are \( \{4, 5, 6, 7, 8, 9, 10\} \) (7 numbers): \[ \text{Ways for } A = \binom{7}{2} = 21 \]
Step 2: Finding Event B (Maximum number is 7).
If the maximum number must be 7, then 7 is fixed as one of the chosen numbers. The remaining 2 numbers must be chosen from the values strictly less than 7, which are \( \{1, 2, 3, 4, 5, 6\} \) (6 numbers): \[ \text{Ways for } B = \binom{6}{2} = 15 \]
Step 3: Finding Event \( A \cap B \) (Minimum is 3 AND Maximum is 7).
If the minimum is 3 and the maximum is 7, both 3 and 7 are fixed. The remaining 1 number must be chosen from the values strictly between 3 and 7, which are \( \{4, 5, 6\} \) (3 numbers): \[ \text{Ways for } A \cap B = \binom{3}{1} = 3 \]
Step 4: Combining everything using the inclusion-exclusion principle.
\[ \text{Total Favorable Ways} = 21 + 15 - 3 = 33 \] \[ \text{Probability} = \frac{33}{120} = \frac{11}{40} \] This matches option (A) exactly.
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