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Anyone here decent at statistics and probability theory?

Let's say I have 47 balls. There is a 70 % chance that each ball is red, and a 30 % chance it is blue. How many blue balls will I have on average? I suppose it is Poisson distributed (?).

No, this is not my home work.

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8 minutes ago, janrichmond said:

35

Those are the red balls, right?

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Hmmm, it's not 35 I don't think. If straight calculation it would be roughly 14 blue balls but that's not correct as each ball has a 70% chance of being red. I would think most likely all balls would be red but that doesn't account for the 30% that could be blue. So maybe 70% of the 14 would still be red? 4 blue balls? Someone please shed some light on this non abstract mind of mine 😊

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1 minute ago, Kwick1 said:

Help with logic? I have it reversed with just over 9 still being Red.

I'm just guessing... i like quizzes • 1
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Statistics? Math?!!!! Good-bye. 😁

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Could it be 'daffodils?'

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17 hours ago, SoulMonster said:

Anyone here decent at statistics and probability theory?

Let's say I have 47 balls. There is a 70 % chance that each ball is red, and a 30 % chance it is blue. How many blue balls will I have on average? I suppose it is Poisson distributed (?).

No, this is not my home work.

You will have the average amount of blue balls.

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14. Probablility is 30%, so 47 x .3

This could be the idiot's answer though.

Edited by aliexpress
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1 hour ago, aliexpress said:

14. Probablility is 30%, so 47 x .3

This could be the idiot's answer though.

That's how I calculated it, too. But I always have this nagging feeling I am doing it wrong Statistics do that to me. Thanks, though.

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20 minutes ago, SoulMonster said:

That's how I calculated it, too. But I always have this nagging feeling I am doing it wrong Statistics do that to me. Thanks, though.

Ha, we're obviously both on the same page. To be honest I'm an accountant and one thing I've learned in my many years is that if you're confident in your numbers people accept them. They can't be bothered googling 'poisson distributed' ( I did but it looked too hard) they'd rather just accept it.

So the answer is 14. Unless anyone can show me a calculation that says it's wrong...

I believe if you pulled the balls from a hat one by one the answer would be different, much like the odds of getting red in roulette

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I think the 30% (14 blue balls) is too simple. The kicker is the "each" ball has a 70% chance of being red and a 30% chance of being blue. So most likely each ball would be red. This has bugged me. My dad is a mathematician and I'm going to send it to him. If he comes back with 14, I'll accept the 14 and blame law school.

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5 hours ago, SoulMonster said:

That's how I calculated it, too. But I always have this nagging feeling I am doing it wrong Statistics do that to me. Thanks, though.

That's how I would calculate it as well.  Not sure how else you would calculate it.  That's how we calculate poker odds, etc.

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4 hours ago, aliexpress said:

I believe if you pulled the balls from a hat one by one the answer would be different, much like the odds of getting red in roulette

The odds would be 2.33 to 1 that you would pull a red ball over a blue ball.

Edited by Kasanova King
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1 hour ago, Kasanova King said:

The odds would be 2.33 to 1 that you would pull a red ball over a blue ball.

But those odds would change with each ball you drew. I’ve a nagging feeling there’s some kind of cumulative probability wotnot going on here. Are you picking all of the balls or just a set amount?

@SoulMonster is that the exact wording of the question?

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Just now, Dazey said:

But those odds would change with each ball you drew. I’ve a nagging feeling there’s some kind of cumulative probability wotnot going on here. Are you picking all of the balls or just a set amount?

@SoulMonster is that the exact wording of the question?

I am picking all of the balls and the outcomes are independent, i.e. the probability of the last ball being blue is 30 % regardless of what I have picked previously.

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4 minutes ago, SoulMonster said:

I am picking all of the balls and the outcomes are independent, i.e. the probability of the last ball being blue is 30 % regardless of what I have picked previously.

So the balls can be either blue or red until they’re out of the box? What the fuck is this madness? Schrodinger’s box of red and blue balls? Share on other sites

This is similar to die rolls. The probability of rolling a 6 (or any other number) when throwing one die, is 1/6 (because only one of the six sides have 6 pits). If you throw two dice, the probability of both being 6, is 1/36 (1/6 x 1/6). In this way I can calculate the probability of, say, 27 balls being red and the rest being blue, but I am not interested in calculating the probability of a specific outcome, I am interested in knowing what the average outcome will be, e.g. assuming the probabilities are Poisson distributed, what the maximum is. So, let's say I did this 1000 times, and counted the number of blue balls each time, what would the average number be?

If you are interested in the details:

We have an internal R&D project going where we order synthetic DNA sequences. When we order them we ask the subcontractor who does the actual synthesis to insert some randomness in the sequences. Usually you would order something like 5'-ATGGCTAGTCGC-3' and so on, which is a perfectly defined sequence with no room for deviation. But in this project we want the sequences to contain random changes (or mutations), so instead we order something like 5'-ATG11111111TCGC-3' where each 1 in the sequence means that there should be a 70 % of no change and a 30 % chance of a different nucleotide being inserted. We then get millions of these molecules and want to know what the average number of mutations per sequence is.

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23 minutes ago, Dazey said:

But those odds would change with each ball you drew. I’ve a nagging feeling there’s some kind of cumulative probability wotnot going on here. Are you picking all of the balls or just a set amount?

Yep, they would. Then you would re-adjust the odds depending on which ball you picked each time. (Unless you put the ball you picked back in.  If so, the odds would still be 2.33 to 1)

Edited by Kasanova King
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See!

(This is why I took up history in high-school.)

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15 minutes ago, SoulMonster said:

This is similar to die rolls. The probability of rolling a 6 (or any other number) when throwing one die, is 1/6 (because only one of the six sides have 6 pits). If you throw two dice, the probability of both being 6, is 1/36 (1/6 x 1/6). In this way I can calculate the probability of, say, 27 balls being red and the rest being blue, but I am not interested in calculating the probability of a specific outcome, I am interested in knowing what the average outcome will be, e.g. assuming the probabilities are Poisson distributed, what the maximum is. So, let's say I did this 1000 times, and counted the number of blue balls each time, what would the average number be?

As long as you keep all the balls in, the probability won't change.

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I didn't think this was a which color ball would be pulled from the bag but how many balls would be blue if each ball has a 70% chance of being blue. It ends up being a simple calculation .70(47) & .30(47). So 14 blue balls are in the bag.

@SoulMonster you're correct.

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