Question

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

• A. 11/40
• B. 1/3
• C. 3/10
• D. 7/10

Answer 1

Answer: (A)

11/40

Answer 2

Total ways to choose 3 numbers from 10 is $\binom{10}{3} = 120$. Let A be the event that the minimum is 3. This means 3 is chosen, and 2 numbers are chosen from {4, 5, 6, 7, 8, 9, 10}. So, $|A| = \binom{1}{1} \times \binom{7}{2} = 21$. Let B be the event that the maximum is 7. This means 7 is chosen, and 2 numbers are chosen from {1, 2, 3, 4, 5, 6}. So, $|B| = \binom{1}{1} \times \binom{6}{2} = 15$. Let $A \cap B$ be the event that the minimum is 3 and the maximum is 7. This means 3 and 7 are chosen, and 1 number is chosen from {4, 5, 6}. So, $|A \cap B| = \binom{1}{1} \times \binom{1}{1} \times \binom{3}{1} = 3$. Using the inclusion-exclusion principle, $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{21}{120} + \frac{15}{120} - \frac{3}{120} = \frac{33}{120} = \frac{11}{40}$.