Question

If three numbers are selected from the set of the first $20$ natural numbers, then the probability that they are in G.P is

• A. $\frac{1}{285}$
• B. $\frac{4}{285}$
• C. $\frac{11}{1140}$
• D. $\frac{1}{71}$

Answer 1

Answer: (C)

$\frac{11}{1140}$

$n\left(S\right)=$ Total number of water ways of selecting 3 numbers from first 20 natural numbers

${=}^{20}{C}_3$

$=\frac{20!}{17!3\times 2}=\frac{20\times 19\times 18\times 17!}{17!\times 3\times 2}=1140$

Three numbers are in G.P, the favorable cases are

$1,2,4;1,3,9;1,4,16;$

$2,4,8;2,6,18;$

$3,6,12;$

$4,8,16;4,6,9;$

$5,10,20;$

$8,12,18;$

$9,12,16.$

$\therefore n\left(E\right)=$ The number of favorable cases

$\therefore$ Required probability $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{11}{1140}$