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)
$= \frac{n^{2}}{4}\cdot\frac{3}{n +1}\cdot\frac{2}{n} \cdot\frac{1}{n-1} = \frac{3n}{2\left(n^{2} -1\right)}$