Probability ... tossing coins

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by shankar.ashwin » Mon Nov 07, 2011 5:28 am
5 coins tossed, you have 2^5 outcomes - 32 possible outcomes.

At any of the outcomes you either have higher number of heads (or) tails. (you cannot have possibility of equal heads and tails in odd coins)

Since P(head) = P(tail), we will have 50% cases where n(H) > n(Tails) and the other half where n(T) > n(H).

Hence probability of at least 3 heads will be 1/2

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by user123321 » Mon Nov 07, 2011 9:43 am
if five coins are tossed...then probability of having three coins turning head is..
5c3/32

using same approach you can find for others and sum it up to get.
(5c3 + 5c4 + 5c5)/2^5 = 16/32 = 1/2

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Want to do it right the first time.

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by praveenpurwar2004 » Mon Nov 07, 2011 10:32 am
You can use Binomial Probability for these type of questions.

The probability of getting exactly k successes in n trials is
P= kCn * p^k * (1-p)^(n-k)
where p is the probability of a success, q is 1 − p, or the probability of a failure, and

5c3 * (1/2)^3 * (1/2)^2 + 5c4 * (1/2)^3 * (1/2)^2 + 5c5 * (1/2)^3 * (1/2)^2
= 10/32 + 5/32 + 1/32
= 16/32 = 1/2

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by Sepideh Schoenfeld » Tue Nov 08, 2011 9:49 am
Thank you guys!
I understood it now!