Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition),
Question:

Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is :

1. $\frac{15}{101}$

2. $\frac{5}{101}$

3. $\frac{5}{33}$

4. $\frac{10}{99}$

Correct Option: , 3

Solution:

Out of 11 consecutive natural numbers either 6 even and 5 odd numbers or 5 even and 6 odd numbers

when 3 numbers are selected at random then total cases $={ }^{11} \mathrm{C}_{3}$

Since these 3 numbers are in A.P. Let no’s are $a, b, c$

$2 \mathrm{~b} \Rightarrow$ even number

$a+c \Rightarrow\left(\begin{array}{l}\text { even }+\text { even } \\ \text { odd }+\text { odd }\end{array}\right)$

so favourable cases $={ }^{6} \mathrm{C}_{2}+{ }^{5} \mathrm{C}_{2}$

$=15+10=25$

$\mathrm{P}\left(3\right.$ numbers are in A.P. $\left.=\frac{25}{{ }^{11} \mathrm{C}_{3}}=\frac{25}{165}=\frac{5}{33}\right)$