Question

If three numbers be selected randomly from a set of numbers $\left(3^0,3^1,3^2,...,3^n\right)$without replacement, then the probability that selection of numbers form an increasing geometric sequence is

• A. $\frac{3}{2n}$ if $n$ is even
• B. $\frac{3}{2n}$ if $n$ is odd
• C. $\frac{3n}{2\left(n^2-1\right)}$ if $n$ is odd
• D. $\frac{3n}{2\left(n^2-1\right)}$ if $n$ is even

Answer 1

Answer: (B), (D)

$n\left(S\right)=$ number of ways of selection of $3$ numbers out of $\left(n+1\right)$ numbers $={}^{n+1}{C}_3$
A = Event of selecting $3$ numbers when $n$ is odd
B = Event of selecting $3$ numbers when $n$ is even
Now, number of triplets of the form $3^r,3^{r+2}{3}^{r+2}$
$\left(0\le r\le n-2\right)=n-1$
Number of triplets of the form $\left(3^r,3^{r+2},3^{r+4}\right),0\le r\le n-4=n-3$
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Number of triplets of the form$\left(3^r{3}^{r+\frac{n-1}{2}},3^{r+\left(n-1\right)}\right)$ when $n$ is odd i.e., we have the set $\left(1,3,3^2,3^3\right)$ and we have the triplets $\left(1,3,3^2\right)$ and $\left(3,3^2,3^3\right)=2$
$\therefore n\left(A\right)=$ When $n$ is odd the total number of favourable cases
$=\left(n-1\right)+\left(n-3\right)+...+4+2=\frac{n^2-1}{4}$
and number of triplets of the form$\left(3^r{3}^{r+\frac{n}{2}}{3}^{r+n}\right)$ when $n$ is even i.e., we have only the set $1,3,3^2$ i.e. only $1$.
$n\left(B\right)$ = When n is even, the total number of favourable outcomes
$=\left(n-1\right)+\left(n-3\right)+...+3+1=\frac{n^2}{4}$
$P\left(A\right)$ = Probability (when n is odd)
$=\frac{\frac{n^2-1}{4}}{{}^{n+1}{C}_3}=\frac{n^2-1}{4}\times \frac{3}{n+1}\times \frac{2}{n}\times \frac{1}{n-1}=\frac{3}{2n}$
$P\left(B\right)$ = Probability (when n is even)
$=\frac{n^2}{4}\cdot \frac{3}{n+1}\cdot \frac{2}{n}\cdot \frac{1}{n-1}=\frac{3n}{2\left(n^2-1\right)}$