IIT JEE , AIEEE Forum - Mathematics , Algebra

raja987654321

Three distinct numbers are randomly selected from first 20 natural numbers .Find the probability that these numbers are in a geometric progression?

venkat_tatolu

Required probability =4c3/20c3

=1/285 (numerator is because of 2,4,8,16 which are in GP.

krishna.gopal

A G.P of ratio 2 will be a*(1,2,4) and there are 5 such series
A G.P of ratio 3 will be a*(1,3,9) and there are 2 such series
A G.P of ratio 4 will be a*(1,4,16) and there are 1 such series

Another point that if 1,2, 4 is in G P then 4,2,1 is also in G P but 2,1,4 is not a GP.
So total numbers of permutations which form GP are = 2*(1+2+5)=16
Toatl number of permutations of choosing 3 numbers = 20*19*18

So answer = 16/(20*19*18)

MOHD

there is only one way of having three nos in gp thus ans=1/20

rajatsen91

total number of selections = 20C3 = 57*20
the numbers which are in g.p are
1,2,4,8,16
2,6,12
3,6,12
1,3,9
2,10,20
1,4,16
so probability = 6C1*5C3*3C3*3C3*3C3*3C3*3C3/(57*20) = 60/(57*20) = 1/19
RATE ME

raja987654321

sorry guys none of you matched the answer which turns out to be 11/1140

nadeemoidu

Total no. of ways of ways of selecting 3 nos. =20C3= 1140

Possible GPs
1 2 4
1 3 9
1 4 16
2 4 8
2 6 18
3 6 12
4 8 16
5 10 20

Now the ones that everyone missed
4 6 9
8 12 18
9 12 16

Who said that to get 3 integers in a GP , the common difference should an integer. :)
so finally the answer is 11/1140

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