Question

If $12$ tickets numbered $0,1,2,....11$ are placed in a bag, and three are drawn out, then the chance that the sum of the numbers on them is equal is $12$ is

• A. $\frac{2}{5}$
• B. $\frac{1}{5}$
• C. $\frac{3}{59}$
• D. $\frac{3}{55}$

Answer 1

Answer: (A)

$\frac{2}{5}$

Given total no. of tickets $=12$

Three tickets are selected from a bay in ${}^{12}{C}_3$ ways.

The sum of three tickets should be $12,$ there are $18$ ways by which sum of the tickets choosen can be $12$

$1).6,3,3=3!/2!=3$

$2).7,3,2=3!=6$

$3).8,2,2=3!/2!=3$

$4).5,5,2=3!/2!=3$

$5).4,4,4=3!/3!=1$

$6).5,6,1=3!=6$

$7).10,1,1=3!/2!=3$

$8).9,2,1=3!=6$

$9).8,1,3=3!=6$

$10).7,4,1=3!=6$

$11).6,4,2=3!=6$

$12).6,6,0=3!/2!=3$

$13).5,3,4=3!=6$

$14).5,7,0=3!$

$15).8,4,0=3!$

$16).9,3,0=3!$

$17).10,2,0=3!$

$18).11,1,0=3!$

$\Rightarrow \frac{Sumofthetotalno.ofways}{Totaloutcomes}=P$

$\Rightarrow P=\frac{88}{{}^{12}{C}_3}=\frac{2}{5}$

Hence, the answer is $\frac{2}{5}.$