Question

Out of $21$ tickets marked with numbers from $1$to $21$, three are drawn at random. The chance that the numbers on then are in AP is:

• A. $\frac{10}{133}$
• B. $\frac{9}{133}$
• C. $\frac{9}{1330}$
• D. None of these

Answer 1

The correct option is A

$\frac{10}{133}$

Explanation for correct option

Calculating the probability of selecting three tickets that are in A.P:

Given, total number of tickets = $21$

Total number of ways of selecting three tickets at random$={}^{21}C_3$

If the common difference is $1$, then the possible ways$=1,2,3;4,5,6;\dots 19,20,21=19$ ways

If the common difference is $2$, then the possible ways$=1,3,5;2,4,6;\dots 17,19,21=17$ways

Similarly, in the same way, if the common difference is $10$, then the possible ways$=1,11,21=1$way

So, if the common difference is greater than $10$, then there will be no favourable case.

Hence, the total number of favourable cases:

$$\begin{gathered} =19+17+15+13+11+9+7+5+3+1 \\ =100 \end{gathered}$$

Therefore, the required probability:

$$\begin{gathered} =\frac{100}{{}^{21}C_3} \\ =\frac{100}{1330} \\ =\frac{10}{133} \end{gathered}$$

Hence, the correct answer is Option (A).