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04 Sep 2005, 16:41
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Hello Everyone
here is a problem on tossed coins from this software
when n coins are tossed simultaneously in how manyof the outcomes will one get all heads or all tails
plz explains yours works
thanks
here is a problem on tossed coins from this software
when n coins are tossed simultaneously in how manyof the outcomes will one get all heads or all tails
plz explains yours works
thanks
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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04 Sep 2005, 21:27
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I believe it is 1/2^n. I don't think it matters whether the coins are tossed simultaneously or individually.
05 Sep 2005, 21:29
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U are right. This is a unique case of Binomial Distn. nCr.p^r.(1-p)^ (n-r)
here n = total number of tosses [here n]
r = no. of times desired outcome is possible [here = n , all heads]
p = probability of the desired outcome [here 1/2 = p of head in each toss]
so nCn.(1/2)^n.(1-1/2)^(n-n) = (1/2)^n.
In much more simple terms - p(first coin head).p(2nd coin head).....p(nth coin head) = 1/2.1/2....n times = (1/2)^n
However it is always better to stick to the general concept as that can be used more smoothly in all cases. For Eg.
Whats the P of getting 4 sixers on rolling 6 dices.
using above funda.
n = 6
r = 4
p = 1/6
P = 6C4.(1/6)^4.(5/6)^2.
Another important property is
SIGMA[r = 0 to n] nCr.p^r.(1-p)^ (n-r) = 1
Thats the sum of all cases and hence probab is 1.
Please let me know if you have any comments.
here n = total number of tosses [here n]
r = no. of times desired outcome is possible [here = n , all heads]
p = probability of the desired outcome [here 1/2 = p of head in each toss]
so nCn.(1/2)^n.(1-1/2)^(n-n) = (1/2)^n.
In much more simple terms - p(first coin head).p(2nd coin head).....p(nth coin head) = 1/2.1/2....n times = (1/2)^n
However it is always better to stick to the general concept as that can be used more smoothly in all cases. For Eg.
Whats the P of getting 4 sixers on rolling 6 dices.
using above funda.
n = 6
r = 4
p = 1/6
P = 6C4.(1/6)^4.(5/6)^2.
Another important property is
SIGMA[r = 0 to n] nCr.p^r.(1-p)^ (n-r) = 1
Thats the sum of all cases and hence probab is 1.
Please let me know if you have any comments.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
