Re: Attention mathematicians: A probability problem

On Nov 3, 11:07 am, "jerry_fried...@xxxxxxxxx"
<jerry_fried...@xxxxxxxxx> wrote:
On Nov 3, 9:47 am, Bob G <mrbobja...@xxxxxxxxx> wrote:

Obama is leading in five battleground states, but only within the
margins of error.

What is the probability that he will win at least one of them?

At least two?

That he will lose all?

Assume a 4% margin of error, if that's relevant.

You have to know a couple more things (where "a couple" can be three).

Do the pollsters know what they're doing?

If so, what are Obama's leads in those states?  For example, if he
leads by 0.5% in some state, his probability of winning it is less
than if he leads by 3.5%.

What are the distributions of errors in these polls?  That's needed to
calculate the probability of winning from the difference in
percentages.  If the distribution is normal, you use some kind of
error function, but I forget. You also need to know the meaning of
"margin of error".  According to
<http://en.wikipedia.org/wiki/Opinion_poll>, it's usually a 95%
confidence interval.

If you know the probabilities of Obama winning each state are P1, P2,
etc., the probability that Obama will lose all five of those states is

(1-P1)(1-P2)(1-P3)(1-P4)(1-P5).

The probability that he'll win at least one is 1 - (1-P1)(1-P2)(1-P3)
(1-P4)(1-P5).

Okay. Having the formula here helps, because I didn't answer that
question. At least one is 1-1/32 or 31/32.

The probability that he'll win at least two is

1 - (1-P1)(1-P2)(1-P3)(1-P4)(1-P5) - P1(1-P2)(1-P3)(1-P4)(1-P5) - (1-
P1)P2(1-P3)(1-P4)(1-P5) - three more terms following that pattern.

That works out to: 31/32 - 1/32 - 1/32 - 1/32 -1/32 - 1/32 = 26/32.

Unless I've done something I should know better than.

Just saved me a lot of time writing the right formula for at least two
and at least one. P1=1-P1 for all P when P=50%.

--
Jerry Friedman voted on Saturday.  Isn't this over yet?

.

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