Question

Out of the first $20$ consecutive natural numbers, $3$ numbers are chosen at random. If these $3$ numbers are in arithmetic progression with common difference $d\in\mathbb N$, then the probability of getting those $3$ numbers whose common difference is a prime number is

• A. $\dfrac{4}{9}$
• B. $\dfrac{1}{3}$
• C. $\dfrac{23}{45}$
• D. $\dfrac{29}{90}$

Hint:

For a three-term A.P., count possible starting terms using the condition $a+2d\le n$.

Answer 1

The Correct Option is C

Step 1: Set up a 3-term AP.
A three-term AP from $\{1,\dots,20\}$ has the form $a,\ a+d,\ a+2d$ with common difference $d\in\mathbb N.$ The largest term must satisfy $a+2d\le20.$
Step 2: Count APs for each $d$.
For a fixed $d$, $a$ runs from $1$ to $20-2d$, giving $20-2d$ APs. The value $d$ can be $1$ up to $9$ (since $2d<20$).
Step 3: Total number of APs.
$\sum_{d=1}^{9}(20-2d)=18+16+14+12+10+8+6+4+2=90.$ So there are $90$ equally likely APs.
Step 4: Pick the prime differences.
Primes among $1,\dots,9$ are $2,3,5,7.$ The matching counts of APs are $20-4=16,\ 20-6=14,\ 20-10=10,\ 20-14=6.$
Step 5: Add the favourable cases.
$16+14+10+6=46.$
Step 6: Form the probability.
$P=\dfrac{46}{90}=\dfrac{23}{45}.$ \[ \boxed{\dfrac{23}{45}} \]