Four numbers chosen at random (no replacement) from 1,,20.

Question

Four numbers chosen at random (no replacement) from $\{1,\dots,20\}$. Statement-1: probability they form an A.P. (in some order) is $\dfrac{1}{85}$. Statement-2: possible common differences are $\pm1,\dots,\pm5$.

Accepted Answer

Counting A.P.s by common difference $d=1,\dots,6$: $17,14,11,8,5,2$ respectively (up to $d=6$, since $d$ can be as large as 6 for a 4-term A.P. within 1--20).

$$P = \frac{17+14+11+8+5+2}{{}^{20}C_4} = \frac{57}{4845}=\frac{1}{85}$$

Statement-2 is false since $d$ can be 6 as well (Statement-1 alone is true, so answer B).

Solutions · 1