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A coin has probability p of landing heads. 100 such coins are flipped and afterwards, every coin that landed tails is flipped again. What is the variance of the number of heads?

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Got something from the answer below? –  Did Feb 9 '13 at 10:02
    
the distribution is binomial, and the variance is $nq(1-q)= n(2p-p^2)(1-2p+p^2)$ –  user59036 Feb 10 '13 at 20:05
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Hint: Each coin has the same probability $q$ to show heads at the end of the procedure, independently of the others. This may happen either because it first showed heads, or because it first showed tails, was flipped again, and then showed heads. Hence $q=$ $________$. There are $n=100$ coins. Thus the distribution of the number of heads is $____________$, whose variance is $nq(1-q)=$ $__________$.

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since each coin first showed heads, or first showed tails, flipped again, and then showed heads, q = q + (1-q)q. Am I on the right track? –  user59036 Feb 7 '13 at 6:31
    
On the right track, yes, except that you are confusing p and q. –  Did Feb 7 '13 at 8:59
    
In the problem it said the probability of a coin landing on head is p, I don't quite understand why you are using q in the hint? –  user59036 Feb 7 '13 at 18:36
    
Read: probability q to show heads at the end of the procedure. –  Did Feb 7 '13 at 18:37
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use the binomial variance which is np(1-p)

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