2. ## Re: probability

Hello, Trefoil2727!

In an election, there are 3 candidates.
The results are in the ratio-$\displaystyle A:B:C \:=\:5:3:2.$
Three votes are chosen at random.
Find the probability that exactly 2 of them voted for the same candidate.

I got:-$\displaystyle (\tfrac{5}{10})^2(\tfrac{3}{10}) + (\tfrac{5}{10})^2(\tfrac{2}{10}) + (\tfrac{3}{10})^2(\tfrac{5}{10}) + (\tfrac{3}{10})^2(\tfrac{2}{10}) + (\tfrac{2}{10})^2(\tfrac{5}{10}) + (\tfrac{2}{10})^2(\tfrac{3}{10}) \;=\;\tfrac{11}{50}$

But the answer is:-$\displaystyle \tfrac{33}{50}$

You had the right idea, but you forgot a detail . . .

$\displaystyle P(A,A,B) \;=\; \underbrace{{3\choose2}}_{\text{This!}}\left(\frac {5}{10}\right)^2\left( \frac{3}{10}\right)$

3. ## Re: probability

Originally Posted by Soroban
Hello, Trefoil2727!

You had the right idea, but you forgot a detail . . .

$\displaystyle P(A,A,B) \;=\; \underbrace{{3\choose2}}_{\text{This!}}\left(\frac {5}{10}\right)^2\left( \frac{3}{10}\right)$
but why?

4. ## Re: probability

Originally Posted by Soroban
Hello, Trefoil2727!

You had the right idea, but you forgot a detail . . .

$\displaystyle P(A,A,B) \;=\; \underbrace{{3\choose2}}_{\text{This!}}\left(\frac {5}{10}\right)^2\left( \frac{3}{10}\right)$
Doesn't the above assume(incorrectly) that after you remove an A-vote then the probability remains the same(5/10) for getting another A-vote?
I mean it's not a 10-sides-dice that is thrown 3 times for the probability to stay the same. After you remove one A-vote(the one 5/10 of the (5/10)^2) the probability to get another A-vote should lower a bit from 5/10.

My way of calculating the desired probability results in that the result depends on the total numbers of votes, let's say total number of votes it is 10N(we don't choose N, for the sake of simplicity to not have to deal with fractions like 5N/10, but it's equivalent whatever we choose). And it is:

Which is equal to:

$\displaystyle \frac{{6{N^2}\left( {25(5N - 1) + 21(3N - 1) + 16(2N - 1)} \right)}}{{20N(10N - 1)(10N - 2)}} =$

$\displaystyle \frac{{3N\left( {25(5N - 1) + 21(3N - 1) + 16(2N - 1)} \right)}}{{10(10N - 1)(10N - 2)}} =$

$\displaystyle \frac{{3N\left( {125N - 25 + 63N - 21 + 32N - 16} \right)}}{{10(10N - 1)(10N - 2)}} =$

$\displaystyle \frac{{3N\left( {220N - 62} \right)}}{{10(10N - 1)(10N - 2)}} =$

$\displaystyle \frac{{3N\left( {110N - 31} \right)}}{{5(10N - 1)(10N - 2)}}$

That can't be simplified any more and it depends on the value of N. The value of the votes. And it is different if we talk about 10 voters or 100.
Am i mistaken?

5. ## Re: probability

Originally Posted by ChessTal
Doesn't the above assume(incorrectly) that after you remove an A-vote then the probability remains the same(5/10) for getting another A-vote?
I mean it's not a 10-sides-dice that is thrown 3 times for the probability to stay the same. After you remove one A-vote(the one 5/10 of the (5/10)^2) the probability to get another A-vote should lower a bit from 5/10.

My way of calculating the desired probability results in that the result depends on the total numbers of votes, let's say total number of votes it is 10N(we don't choose N, for the sake of simplicity to not have to deal with fractions like 5N/10, but it's equivalent whatever we choose). And it is:

Which is equal to:

$\displaystyle \frac{{6{N^2}\left( {25(5N - 1) + 21(3N - 1) + 16(2N - 1)} \right)}}{{20N(10N - 1)(10N - 2)}} =$

$\displaystyle \frac{{3N\left( {25(5N - 1) + 21(3N - 1) + 16(2N - 1)} \right)}}{{10(10N - 1)(10N - 2)}} =$

$\displaystyle \frac{{3N\left( {125N - 25 + 63N - 21 + 32N - 16} \right)}}{{10(10N - 1)(10N - 2)}} =$

$\displaystyle \frac{{3N\left( {220N - 62} \right)}}{{10(10N - 1)(10N - 2)}} =$

$\displaystyle \frac{{3N\left( {110N - 31} \right)}}{{5(10N - 1)(10N - 2)}}$

That can't be simplified any more and it depends on the value of N. The value of the votes. And it is different if we talk about 10 voters or 100.
Am i mistaken?
Since you are not given the number of voters, you should assume it is large. In this context large means treat it as though it is infinite, which gives you the same results as sampling with replacement for this type of problem.

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6. ## Re: probability

Originally Posted by zzephod
In this context large means treat it as though it is infinite, which gives you the same results as sampling with replacement for this type of problem.

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Obviously, but i don't find quite logical the idea of an infinite number of voters. If i would be given the problem in its initial form and have been told later on that i should have assumed infinite number of voters so the 5:3:2 ratio ALWAYS holds, then i would kill the problem author.

7. ## Re: probability

Thanks for this solution.. It build my concept of probability..