Question

Out of $21$ tickets marked with numbers $1$ to $21$, $3$ tickets are drawn at random. Find the probability that the $3$ numbers on them are in A.P.

• A. $10/133$
• B. $101/1330$
• C. $99/1330$
• D. $11/133$

Answer 1

The correct option is A $10/133$

Number of tickets $=21$
Number of ways of drawing three tickets ${=}^{21}{C}_3=1330$
Now we shall find the number of favorable cases.
If common difference is $1$, the favorable group of number are $1,2,3;2,3,4;3,4,5;...;19,20,21.$
These groups are $19$ in numbers.
If common difference is $2$, the favorable groups of number are $1,3,5;2,3,6;3,5,7;...;17,19,21.$
These groups are $17$ in number.
If common difference is $3$, then favorable groups of number are $1,4,7;2,5,8;3,6,9;...;15,18,21.$
These groups are $15$ in numbers.
If common difference is $10$, the favorable groups of number is $1,11,21.$
Thus number of favorable cases $=19+17+15+....+1$
$=\frac{10}{2}\{2\left(19\right)+(10-1)(-2)\}=5(38-18)=100$
Therefore $P$(numbers are in A.P) $=\frac{100}{1330}=\frac{10}{133}$