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  #21  
Old 09-02-2012, 08:05 AM
coolguy coolguy is offline
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Default Re: question about probability

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Originally Posted by yaser View Post
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.
WOW!Thanks for the detailed explanation!It was over my head for a long time.
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  #22  
Old 08-28-2013, 03:32 AM
weehong weehong is offline
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Default Re: question about probability

Quote:
Originally Posted by yaser View Post
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.
I captured the above idea, however I am confused why the following approach gives wrong answer:

Probability of all heads in 10 flips by one coin: p = 0.5^10
Probability of all heads in 10 flips by any of 1000 coin: 1000*p = 97.7%

In above I assumed there were 1000 10-flips.
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  #23  
Old 08-30-2013, 02:02 AM
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yaser yaser is offline
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Default Re: question about probability

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Originally Posted by weehong View Post
I captured the above idea, however I am confused why the following approach gives wrong answer:

Probability of all heads in 10 flips by one coin: p = 0.5^10
Probability of all heads in 10 flips by any of 1000 coin: 1000*p = 97.7%

In above I assumed there were 1000 10-flips.
In order to multiply the probabilities in the "all" case, you need the events to be independent, and we have that for the coin flips. In order to add the probabilities in the "any of" case, you need the events to be disjoint, i.e., they cannot simultaneously occur. The first coin giving 10 heads is not disjoint from the second coin giving 10 heads, so when you add their probabilities you are double counting the overlap which is "both coins giving 10 heads each."
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  #24  
Old 09-01-2013, 12:41 AM
weehong weehong is offline
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Default Re: question about probability

Thank you very much, Professor. The explanation is spot on, very helpful.
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