Three numbers are chosen at random from the first 20 numbers 1, 2, 3, … 20. The probability the selectednumbers forms a G.P., is

Three numbers can be chosen out of 20 in 20C3ways.For different values of common ratios, G.P.’s arelisted below.r G.P. Number of G.P.’s2(1,2,4),(2,4,8),(3,6,12)(4,8,16),(5,10,20)53(1,3,9),(2,6,18)24(1,4,16)132(4,6,9),(8,12,18)243(9,12,16)1 Total11∴ probability of required event =11 20C3

Three numbers are chosen at random from the first 20 numbers 1, 2, 3, … 20. The probability the selectednumbers forms a G.P., is

Three numbers can be chosen out of 20 in 20C3ways.For different values of common ratios, G.P.’s arelisted below.r G.P. Number of G.P.’s2(1,2,4),(2,4,8),(3,6,12)(4,8,16),(5,10,20)53(1,3,9),(2,6,18)24(1,4,16)132(4,6,9),(8,12,18)243(9,12,16)1 Total11∴ probability of required event =11 20C3

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Three numbers are chosen at random from the first $20$ numbers $1, 2, 3, \dots 20 .$ The probability the selected
numbers forms a $G . P .$ , is

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$1 / 20$

$1 / 30$

$1 / 45$

$\frac{11}{^{20} C_{3}}$

answer is D.

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Detailed Solution

Three numbers can be chosen out of $20$ in $^{20} C_{3}$

ways.
For different values of common ratios, $G . P .$ ’s are
listed below.

$r$	$G.P.$	$N u m b e r o f G . P . ’ s$
2	$\begin{aligned} (1, 2, 4), (2, 4, 8), (3, 6, 12) \\ (4, 8, 16), (5, 10, 20) \end{aligned}$	5
3	$(1, 3, 9), (2, 6, 18)$	2
4	$(1, 4, 16)$	1
$\frac{3}{2}$	$(4, 6, 9), (8, 12, 18)$	2
$\frac{4}{3}$	$(9, 12, 16)$	1
	Total	11