Question

If $3$ numbers are selected from the first 15 natural numbers, then the probability that the numbers are in A.P is

• A. $\frac{7}{65}$
• B. $\frac{9}{65}$
• C. $\frac{8}{65}$
• D. $\frac{6}{65}$

Answer 1

The correct option is A $\frac{7}{65}$

Mentioned Numbers $1,2,3...15$

i) Taking common difference $=1$

Sequence will be :- $(1,2,3),(2,3,4).....(13,14,15)$

Thus the no. of such triplets found would be $=13$................$\left(a\right)$

ii) Taking common difference $=2$

Sequence will be:- $(1,3,5)(2,4,6)$$...............(11,13,15)$

Thus the no. of such triplets found would be $=11$................$\left(b\right)$

iii) Taking common difference $=3$

Sequence will be $(1,4,7)$ $(2,5,8)$ $.............(9,12,15)$

Thus the no. of such triplets found would be $=9$.................$\left(c\right)$

Hence, the final triplet would be: $(1,8,15)$

Therefore the common difference of the total number of pairs formed would get eventually reduced form $(13,11,\mathrm{9......1})$

Therefore the no. of such cases are

$\left(a\right)+\left(b\right)+\left(c\right)......$

$=13+11+9+...........+1$

$=49$

Total no. of possible triplets from $15$ natural nos

${=}^{15}{C}_3$

$=\frac{15\times 14\times 13}{3\times 2\times 1}$

$=35\times 13$

So, required probability $=\frac{49}{35\times 13}$

$=\frac{7}{65}$