Question

Three numbers are chosen at random withoutreplacement from the set of integers $1,2,3,....10$. The probability that the minimum of the chosen numbers is $3$ or the maximum of the chosen numbers is $7$, is equal to

• A. $23/120$
• B. $13/120$
• C. $13/60$
• D. $11/40$

Answer 1

The correct option is D $11/40$

Case 1: When 3 is minimum
Then the number of choices for choosing 2 cards other then $\mathrm{7....4},5,6,8,9,10=6C_2=15$
The number of ways when 7 is max but 3 not included: $1,2,4,5,6=10$ ways
When both 3,7 are present $8$ ways
Total desired cases $33$
Total ways of picking 3 numbers at random $=10C_3=120$
Probability = $33/120=11/40$