I have a statistics problem that is probably simple when you know the answer, but … I don’t.

I have a bag with 7 white balls and one black ball. I draw a ball from it at random and then put it back in the bag. I do this a total of 8 times. What is the probability that I draw the black ball exactly once?

Thanks for saving my sanity!!

114%

That’s no better than my answer DD 🙂

100%- basically you should draw each ball once in a random trial

That said this wont happen and what will happen is that in say 8000 picks you will pick it 1000 times and sometimes you will pick it once and sometimes more than this and sometimes you wont pick it

I think its going to be one of those that depends on exactly how you word the question

You guys are hopeless !! 🙂

4.9%

I reckon it’s 1/8 * (7/8)^7

Approx 5%…

Edit: See what Stoner said above…

says the one who has to ask the internet 😉

39% for all 8 ways of doing it.
4.9% for it drawing black the first time.

I reckon it’s 1/8 * (7/8)^7

I though that but isn’t it the prob of black on the first draw and then white?

quite right. Stoat’s on the money, I forgot to *8 for all the permutations.

If it’s any defence, my scribblings here do have the *8 on it.

Although… on second thoughts…

If I draw the balls in order, then I can get (b for black, w for white)

bwwwwwww
wbwwwwww
wwbwwwww
wwwbwwww
wwwwbwww
wwwwwbww
wwwwwwbw
wwwwwwwb

Now, each of those probabilities is (1/8) * (7/8)^7… but there are 8 of them..

So, to revise my original answer… about 40%

I am possibly very wrong though…

Part 2

If there are 4 black balls and 4 white, what is the prob of drawing 4 white and 4 black.

Now, each of those probabilities is (1/8) * (7/8)^7… but there are 8 of them..

So, to revise my original answer… about 40%

I followed that logic but doesn’t it seem odd to you that this rather special outcome would be almost 50:50?

Part 2

If there are 4 black balls and 4 white, what is the prob of drawing 4 white and 4 black.

(0.5)^8 x 64 = 0.25

27%
70 combinations

But this time, that’s cheating with an online combinations calculator and probable BS
36 yrs since a level. Too old.

The thing is, it’s not that special – actually picking the black ball once is the “expected” outcome, ie the one you’d expect to happen most often.

Far too easy a problem, can’t we have something more difficult, like the probability of getting a clear winner from several candidates in an election? 😉

Seems right – without thinking about it too hard I’d expect the formula to be:
(4/8)^8 * 8! / (4! * 4!)
the latter part gives 70 combinations.

Similarly for the first question, number of combinations = 8! / (7! * 1!)

Hopeless 😀

OK smart guys …

3 red balls, one black, 4 white

Prob of 3 red balls and 5 white
Prob of 1 black 7 white

🙂

Your talking about probability, not statistics. Just sayin!

The thing is, it’s not that special – actually picking the black ball once is the “expected” outcome, ie the one you’d expect to happen most often.

Yeah – just checked with 2000 tries on Excel and it’s about right. Just felt odd ….

Your talking about probability, not statistics. Just sayin!

Same difference 🙂

23%

and a half

It’s easier to see how it works if you just use 3 balls… 2 white 1 black..

www = (2/3)^3 = 29.6%
wwb = (2/3)^2 * 1/3 = 14.8%
wbw = 2/3 * 1/3 * 2/3 = 14.8%
wbb = 2/3 * (1/3)^2 = 7.4%
bww = 14.8%
bwb = 7.4%
bbw = 7.4%
bbb = 3.7%

So, for all white the probability is 29.8%, however for 2 white and 1 black it’s approx 45%…

What’s a half between friends.

My brain hurts. If you’re wondering, the real problem is about drilling oil wells. If you plan to drill 8 wells, what is the chance you drill exactly one good one (black ball)

So are the red balls oil wells which explode?

a) (3/8)^3 * (4/8)^5 * 8! / (3! * 5!) = 9.2%
b) 1/8 * (4/8)^7 * 8! / (7! * 1!) = 0.78%

off the top of my head – I think that’s right, but only about 95% certain 😉

Red balls explode your career

Dr J,

It was too late at night to do it the hardway so I just wrote half a dozen lines of code to simulate it and test it and the answer is about 39% of the time.

Now I’m not sure your drilling oil wells is the same? Is there a 1/8 chance of hitting the sweet spot? would you keep going to eight wells if you hit one of the early ones? it wouldn’t be a bad thing if you got more than 1 “hit”.

Depends if there’s a Paveway with your name on it

says the one who has to ask the internet

Says the one who clearly has no clue but thought he’d post an answer anyway.

My brain hurts. If you’re wondering, the real problem is about drilling oil wells. If you plan to drill 8 wells, what is the chance you drill exactly one good one (black ball)

Given that fact that you will have gotten your information from a geologist who in reality won’t actually have a clue (despite what they might tell you) then there is no way of telling.

[pedantic semantics]
This is probability not statistics 😉
[/pedantic semantics]

Given that fact that you will have gotten your information from a geologist who in reality won’t actually have a clue (despite what they might tell you) then there is no way of telling.

True dat 🙂

aracer – you’re my hero 🙂

Yeah, but given input from a geologist, the results bear a lot of similarity with statistics.

I’ll give you the correct answer, guaranteed, if you PPG me a tenner. I used to do these sort of things for fun (back when I had no bills to pay).