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
$=\frac{21+15-3}{120}=\frac{33}{120}=\frac{11}{40}$