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:

• A. $\frac{10}{99}$
• B. $\frac{5}{33}$
• C. $\frac{15}{101}$
• D. $\frac{5}{101}$

Answer 1

The correct option is B $\frac{5}{33}$

Case $:1$
$E,O,E,O,E,O,E,O,E,O,E$
$2b=a+c\to$ Even
$\Rightarrow$ Both $a$ and $c$ should be either even or odd.
$P=\frac{{}^6{C}_2{+}^5{C}_2}{{}^{11}{C}_3}=\frac{5}{33}$
Case $:2$
$O,E,O,E,O,E,O,E,O,E,O$
$P=\frac{{}^5{C}_2{+}^6{C}_2}{{}^{11}{C}_3}=\frac{5}{33}$
Total probability $=\frac{1}{2}\times \frac{5}{33}+\frac{1}{2}\times \frac{5}{33}=\frac{5}{33}$