Question

If three numbers are drawn at random successively without replacement from a set $S=\{1,2, ...10\}$, then the probability that the minimum of the chosen numbers is $3$ or their maximum is $7$ .

• A. $\frac{11}{40}$
• B. $\frac{5}{40}$
• C. $\frac{3}{40}$
• D. $\frac{1}{40}$

Answer 1

Answer: (A)

Total number of possible outcomes $={ }^{10} C_{3}$
Let $A$ be the event that minimum of chosen number is 3 and $B$ be the event that maximum of chosen number is 7 . Then,
$n(A)=$ Number of ways of choosing remaining two numbers from the set
$\{4,5,6,7,8,9,10\}={ }^{7} C_{2}=21$
Similarly, $n(B)=$ Number of ways of choosing remaining two numbers from the set
$\{1,2,3,4,5,6\}={ }^{6} C_{2}=15$
and $n(A \cap B)=$ Number of ways of choosing remaining one number from the set
$\{4,5,6\}={ }^{3} C_{1}=3$
Thus, required probability
$=\frac{21+15-3}{120}=\frac{33}{120}=\frac{11}{40}$