question about probability

canon1230 · #1 04-08-2012, 12:32 AM

If an urn contains 100 green or red marbles, and you sample 10 and they are all green, what is the probability that they are all green?

htlin · #2 04-08-2012, 02:30 AM

Quote:

Originally Posted by canon1230

If an urn contains 100 green or red marbles, and you sample 10 and they are all green, what is the probability that they are all green?

The short reply is that probability does not tell us this answer, unless there are more assumptions.

Put it to the simplest case, if you flip a coin and get a head, what is the head probability of the coin? The answer can be "any non-zero number" but there is no further information to pin it down.

Hope this helps.

GraceLAX · #3 04-08-2012, 03:26 AM

Quote:

Originally Posted by canon1230

If an urn contains 100 green or red marbles, and you sample 10 and they are all green, what is the probability that they are all green?

A physics professor friend wrote this up to help clarify a common statistical misconception.
Do these help? Read through the second one for a similar example.

http://badmomgoodmom.blogspot.com/20...rt-one_16.html
http://badmomgoodmom.blogspot.com/20...-part-two.html

canon1230 · #4 04-08-2012, 08:44 AM

Does the Hoeffding Inequality allow us to say something about this probability?

P[|Ein - Eout| > epsilon] <= 2e^(-2 * epslion^2 * N)

Since Ein = 0, N = 10, setting epsilon to 0.5, the inequality gives us:

P[Eout > 0.5] <= 2e^(-5) = 0.013+

This seems to be saying something nontrivial about Eout.

htlin · #5 04-08-2012, 03:01 PM

Quote:

Originally Posted by canon1230

Does the Hoeffding Inequality allow us to say something about this probability?

P[|Ein - Eout| > epsilon] <= 2e^(-2 * epslion^2 * N)

Since Ein = 0, N = 10, setting epsilon to 0.5, the inequality gives us:

P[Eout > 0.5] <= 2e^(-5) = 0.013+

This seems to be saying something nontrivial about Eout.

The

in Hoeffding is subject to the process of generating the sample (i.e. $E_{in}$ ), not the probability on $E_{out}$ . Indeed it tells us something nontrivial (and that's how we use it in the learning context), but it does not answer your original question.

The question that got answered by Hoeffding is roughly

"What is the probability of a big-Eout urn (many red) for generating such an Ein (all green)?"

not

"What is the probability of Eout being small in the first place?"

The answer to the latter question remains unknown, but even so, we know that having a big Eout is unlikely because of Hoeffding.

Hope this helps.

rukacity · #6 04-10-2012, 07:19 AM

I would think this way:

if p is probability of the outcome then for 10 trials there is p^10 probability of getting all favorable outcome.

jsarrett · #7 04-13-2012, 11:00 PM

Hoeffding's inequality has a free parameter in your question, namely $\epsilon$ . It lest you say that since you have 10 samples, if you want to be 80% sure of the distribution of marbles in the jar ( $\epsilon = 0.2$ ), then the jar is at most $E_{out}$ percent different from the sample.

where
$P(|E_{in}-E_{out}| < \epsilon) \le 2e^{-2\epsilon^2N}$

substituting from your example:
$P(|0-E_{out}| < 0.2) \le 2e^{-2(0.2^2)10}$

simplifying:
$P(|0-E_{out}| < 0.2) \le 0.898657928234443$
so we think that with 80% confidence, the jar is at least 10.2% green.

We probably need a bigger N!

pventurelli@gotoibr.com · #8 04-14-2012, 06:24 PM

In the second lecture, the Professor asked a question about flipping a coin:
What is the probability of getting all ten heads if you flip a coin 10 times count the number of heads, and then you repeat the experiment 1000 times.

The answer he gave was 63% -- I would like to know how this was computed.

Any help would be greatly appreciated.

Thanks!

yaser · #9 04-14-2012, 09:11 PM

Quote:

Originally Posted by pventurelli@gotoibr.com

In the second lecture, the Professor asked a question about flipping a coin:
What is the probability of getting all ten heads if you flip a coin 10 times count the number of heads, and then you repeat the experiment 1000 times.

The answer he gave was 63% -- I would like to know how this was computed.

Any help would be greatly appreciated.

Thanks!

The probability of getting 10 heads for one coin is ${1 \over 2} \times {1 \over 2} \times \cdots \times {1 \over 2}$ (10 times) which is aprroximately ${1 \over 1000}$ .

Therefore, the probability of not getting 10 heads for one coin is approximately $(1-{1 \over 1000})$ .

This means that the probability of not getting 10 heads for any of 1000 coins is this number multiplied by itself 1000 times, once for every coin. This probability is therefore $\approx (1-{1 \over 1000})^{1000}$ .

This is approximately ${1 \over e}$ since $\lim_{n\to\infty} (1-{1\over n})^n = {1\over e}$ . Numerically, ${1 \over e}\approx{1\over 2.718}\approx 0.37$ .

Therefore, the probability of this not happening, namely that at least one coin of the 1000 coins will give 10 heads, is 1 minus that. This gives us the answer of approximately 0.63 or 63% that I mentioned in the lecture.

pventurelli@gotoibr.com · #10 04-15-2012, 04:51 PM

Thank you Professor, that was extremely helpful.

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