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paul99 04-14-2013 08:42 PM

Question on Lecture 2
 
Hi,

In reviewing some of the lectures, I have a question about Lecture 2 (slide 14; ~ 47 minute mark), "Coin analogy", how was 63% arrived at as the answer to the second question:

If you toss 1000 fair coins 10 times each, what is the probability that some coin will get 10 heads?

Thanks,

Paul

yaser 04-14-2013 09:10 PM

Re: Question on Lecture 2
 
Quote:

Originally Posted by paul99 (Post 10404)
Hi,

In reviewing some of the lectures, I have a question about Lecture 2 (slide 14; ~ 47 minute mark), "Coin analogy", how was 63% arrived at as the answer to the second question:

If you toss 1000 fair coins 10 times each, what is the probability that some coin will get 10 heads?

Thanks,

Paul

For each coin, probability of getting 10 heads is ({1 \over 2})^{10} which is approximately 0.001

Therefore, for each coin, the probability of not getting 10 heads is 1 minus that, approximately 0.999

The probability that none of the 1000 coins gets 10 heads is that to the power 1000, and we approximate 0.999^{1000} by 1 \over e since the limit of (1-{1 \over N})^N is 1 \over e.

Finally, the probability that at least one coin will get 10 heads is one minus that, which is 1- {1 \over e} and this is approximately 0.63.


PS: You can quote any part of the video lecture by using the lecture tags, in this case [lecture2] tag, as explained in the sticky thread about including a video segment.

paul99 04-15-2013 12:33 PM

Re: Question on Lecture 2
 
Thanks Prof. Yaser and also for reminding me to use the post tags!


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