Probability of forming an increasing Geometric Progression

[Titan97]

Homework Statement

If three number are chosen randomly from the set ${1,3,3^2,...3^n}$ without replacement, then the probability that they form an increasing geometric progression is?
(a) 3/2n if n is odd
(b) 3/2n if n is even
(c)3n/2(n² -1) if n is even
(d) 3n/2(n² -1) if n is odd

Homework Equations

None

The Attempt at a Solution

This problem can be easily solved if I chose the end terms first.
Let $3^a$ be the first term and $3^b$ be the third term. Then the middle term has to be $3^{\frac{a+b}{2}}$.
Now, both $a$ and $b$ have to be even or both has to be odd.
let $n$ be odd. Then number of terms with even exponents=$\frac{n+1}{2}$
number of terms with odd exponents=$\frac{n+1}{2}$
Total number of ways of choosing 2 numbers $3^a$ and $3^b$ = $2\cdot ^{\frac{n+1}{2}}C_2$
Probability is $$P(n)=\frac{2\cdot ^{\frac{n+1}{2}}C_2}{^{n+1}C_3}$$
But I am not getting the answers given in the options.

 

[RUber]

Start by checking n =2 and n = 3, those cases are pretty straightforward and should at least eliminate some of your options.

 

[Titan97]

For n=3, the series is $1,3,9,27$
The possible GPs are : $1,3,9$ and $3,9,27$.
Probability is $2/4=1/2$.
I get the same answer if I substitute $n=3$ in $P(n)$

 

[RUber]

Simplifying your equation for P(n) does give one of the answers for n odd.
Have you worked it out for n even?

$P(n) = 2 \frac{ (\frac{n+1}{2} )! }{ (\frac{n+1}{2}-2 )! 2!} \frac{ (n+1-3)!(3)!}{ (n+1)!}$

 

[Ray Vickson]

For n=3, the series is $1,3,9,27$
The possible GPs are : $1,3,9$ and $3,9,27$.
Probability is $2/4=1/2$.
I get the same answer if I substitute $n=3$ in $P(n)$

I do not get your answer 1/2.

Let's condition on the first chosen number, so $P(GP|i) =$ the (conditional) probability of a GP, given the first chosen number is $i$, for $i = 1, 3$.

Given $i = 1$ as the first number chosen, that leaves $3, 9, 27$. The chance of choosing $3$ next is $1/3$; that leaves $9, 27$, from which we must choose $9$. Thus, $P(GP|1) = (1/3)(1/2) = 1/6$.

Given $i = 3$ that leaves $1, 9, 27$, from which we must next choose $9$, then choose $27$ from the remaining set $1, 27$. Thus, $P(GP|3) = (1/3)(1/2) = 1/6$.

Altogether, we have $P(GP) = P(1) PGP|1) + P(3) P(GP|3) = (1/4)(1/6) + (1/4)(1/6) = 1/12$.

 

[RUber]

I think the question implies that order doesn't matter. Simply that the 3 choices form a geometric progression.
That is to say, for n=2, P=1.

 

[Titan97]

Simplifying your equation for P(n) does give one of the answers for n odd.
Have you worked it out for n even?

$P(n) = 2 \frac{ (\frac{n+1}{2} )! }{ (\frac{n+1}{2}-2 )! 2!} \frac{ (n+1-3)!(3)!}{ (n+1)!}$

Yes. Only difference is that, number of ways of choosing $3^a$ and $3^b$ is $^{\frac{n}{2}+1}C_2+^{\frac{n}{2}}C_2$

 

[Titan97]

@Ray Vickson , are you considering that the probability of choosing the second term changes after choosing the first term? I have considered that all three terms are picked at once.

 

[RUber]

So, are A and C both correct?

 

[Titan97]

Its a "more than one correct" type question.

 

[Ray Vickson]

Its a "more than one correct" type question.

I think the question implies that order doesn't matter. Simply that the 3 choices form a geometric progression.
That is to say, for n=2, P=1.

Well, they DID say increasing GP. I would assume that if they just meant "geometric progression (in any order)" they would have omitted the word "increasing", because it would be redundant.

However, I suppose this is a moot point.

 

[Ray Vickson]

@Ray Vickson , are you considering that the probability of choosing the second term changes after choosing the first term? I have considered that all three terms are picked at once.

Yes, because that is typically what is meant when speaking of choosing things without replacement. However, let's look carefully at the way you wanted to do it.

If order does matter (so {1,3,9} is acceptable, but not {3,1,9} or {9,1,3} ... ) then you need to use permutations, not combinations. The number of permutations of 3 things drawn from 4 distinct things is $_4P_3 = 4 \cdot 3 \cdot 2 = 24$. Among these 24 possibilities, only two of them (namely, {1,3,9} and {3,9,27}) are acceptable, so you end up with a probability of 2/24 = 1/12, exactly as obtained from the sequential, one-at-a-time argument.

If order does not matter we should look at combinations instead of permutations. There are $_4C_3 = 4$ possible "outcomes", among which two of them are "acceptable", so the probability you want would be 1/2.

 

[RUber]

Well, they DID say increasing GP. I would assume that if they just meant "geometric progression (in any order)" they would have omitted the word "increasing", because it would be redundant.

However, I suppose this is a moot if point.

I think the answer selection implied that order didn't matter much more convincingly than the question itself.
My initial sense was to interpret it the same way, but when n=2 plugged into the answer choices gave 3/4 or 1, and 1/6 was nowhere to be seen, I had to infer that order didn't matter.