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What is the probability that 25 out of 2000 women cannot have children if 10 out of 1000 cannot?

2007-11-08 06:15:19 · 3 answers · asked by star_wisher86 1 in Science & Mathematics Mathematics

sorry, it's actually the probability that more than 25 will

2007-11-08 12:09:30 · update #1

3 answers

So 10 per 1000 can't have kids, that makes the chance of any single woman being infertile 1/100.

Are you looking for the chance that exactly 25 out of 2000 won't be able to have children? Not 24, and not 26, but 25?

if that's the case and you have 2000 cases you need a combination of 25 occurrences that have a 1/10 chance and 1975 occurrences that have a 9/10 chance.

If you're looking for likelihoods of getting around that number, or below or above it, you want to look at the binomial distribution (I think).

The math for that if you choose the exact women would be: (.1)^25 * (.9)^1975.

However, you don't care which women in particular, so you multiply by all the possible combinations of 25 women. That's
n!/k!(n-k)! or 2000!/25!(2000-25)!

That number is something like 7.9 times ten to the -59. Or essentially 0.

2007-11-08 07:06:53 · answer #1 · answered by cuharrison 2 · 0 1

The probability for any woman in that group to be sterile is P(S)= 1/100 (I suppose the 10 per 1000 figure is a global statistic).
The probability for exactly 25 women in your 2000 woman group to be sterile is:
P(S)^25*P(notS)^1975*2000C25
Where P(S) is the probability for a woman to be sterile, P(notS) is the probability for a woman to be fertile, and 2000C25 is the number of ways to pick a 25 woman sub-group out of your 2000 women.
P(S) =0.01 P(notS)=0.99 2000C25 = 2000!/(25!*1975!)
The total is P = 0.0446 or 4.46%
The figure is very low because the question is "exactly 25 women out of 2000". If the question is "at least 25 women", you have to compute P(0), P(1)....P(24) and sum it all up. That will give you the probability that fewer than 25 women are sterile in your group. the inverse probability is "at least 25 women being sterile".
so P(at least 25 are sterile) = 1- Σ(i=0 to 24)P(i)
with P(i) = 0.01^i*0.99^(2000-i)*2000!/(i!*(2000-i)!)
The result of that tedious calculation is approximately .156 or 15.6%

2007-11-08 14:41:09 · answer #2 · answered by stym 5 · 0 0

Zero

2007-11-08 14:20:44 · answer #3 · answered by ironduke8159 7 · 0 2

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