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

• A. \( \frac{11}{40} \)
• B. \( \frac{3}{10} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{13}{40} \)

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.

Answer 1

The Correct Option is A

Step 1: Set up the rule.
We use $P(A\cup B) = P(A) + P(B) - P(A\cap B)$. The total ways to pick 3 from 10 is $\binom{10}{3} = 120$.
Step 2: Event A, minimum is 3.
Fix 3, then pick 2 more from $\{4,\ldots,10\}$ (7 numbers): $\binom{7}{2} = 21$.
Step 3: Event B, maximum is 7.
Fix 7, then pick 2 more from $\{1,\ldots,6\}$ (6 numbers): $\binom{6}{2} = 15$.
Step 4: Overlap, min 3 and max 7.
Fix both 3 and 7, then pick 1 from $\{4,5,6\}$: $\binom{3}{1} = 3$.
Step 5: Combine the counts.
Favorable $= 21 + 15 - 3 = 33$.
Step 6: Find the probability.
\[ P = \frac{33}{120} = \frac{11}{40} \] \[ \boxed{\tfrac{11}{40}} \]