Question

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

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

Answer 1

The correct option is B

$\frac{5}{33}$

Explanation for correct option

Calculating the probability of selecting three random numbers that are in A.P

Given, there are $11$consecutive natural numbers

Case-1

$E,O,E,O,E,O,E,O,E,O,E$ (Here, $O$ stands for odd and $E$ stands for even)

For the numbers to be in A.P: $2b=a+c$ (It should be even)

$\Rightarrow$Both $a$ and $c$ should be either even or odd.

In the above case, there are $6$ even numbers and $5$ odd numbers.

Therefore, the probability will be:

$\mathrm{P}=\frac{{}^6\mathrm{C}_2+{}^5\mathrm{C}_2}{{}^{11}\mathrm{C}_3}=\frac{5}{33}$

Case-2

$O,E,O,E,O,E,O,E,O,E,O$

In this case, there are $5$ even numbers and $6$odd numbers.

Hence, the probability will be:

$\mathrm{P}=\frac{{}^5\mathrm{C}_2+{}^6\mathrm{C}_2}{{}^{11}\mathrm{C}_3}=\frac{5}{33}$

Therefore, the Total probability :

$$\begin{gathered} =\left(\frac{1}{2}\right)\times \left(\frac{5}{33}\right)+\left(\frac{1}{2}\right)\times \left(\frac{5}{33}\right) \\ =\frac{5}{33} \end{gathered}$$

Hence, the correct answer is Option $\left(\mathrm{B}\right)$.