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A fair coin is tossed until 100 heads appear. Find the probability that at least 226 tosses will be necessary.

Solution: Let $X$ be the geometric random variable with $p = ½$. Then $S_{100} = X_1 + ... + X_{100}$ is a random variable of the number of coin tosses that result in 100 heads, and $ES_{100} = 100 *(1/p) = 200, Var S_{100} = 100(1-p)/p^2 = 200$.

$$ \begin{align} P(S_{100} \geq 226) &= 1 - P(S_{100} \leq 225) \\[8pt] &\approx 1 - \Phi\left( \frac{225 - 200}{10\sqrt{2}}\right) \\ &= 1 - 0.96145\\ &\approx 0.03855 \end{align} $$

The book's solution is 0.0415. I'm not sure what I'm doing incorrectly.

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  • $\begingroup$ Looks ok to me (though I would use 225.5 - which gives me 0.3568). Are you sure you got all right? (and the book is reliable?) $\endgroup$ – leonbloy Aug 4 '14 at 18:00
  • $\begingroup$ @leonbloy The book is reliable. I have not seen any error in the solution. Is there a different way to solve this problem, so that I can cross check? $\endgroup$ – user90593 Aug 4 '14 at 18:02
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    $\begingroup$ Yes, see here: sosmath.com/CBB/viewtopic.php?t=33427 $\endgroup$ – leonbloy Aug 4 '14 at 18:05
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Your calculation is right, but there is an alternate way of applying the CLT, more direct and -in this case- more precise: Instead of summing geometrical variables (random number of throws for a fixed number or heads: 100), consider the random variable "number of heads" ocurring in a fixed number of throws: 225). This is a binomial variable, and the approximation to a Gaussian will be probably better.

That this approach is valid can be seen by considering that the event "226 or more throws were necessary for getting 100 heads" is equivalent to the event "in 225 throws, less than 100 heads appeared".

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