*Posted by Saketh in : potd , trackback
*
Albert is flipping his lucky coin, keeping track of the cumulative number of heads and tails. If the probability that he reaches $15$ heads before he reaches $14$ tails is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by $1000$.

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**Solution:**
After the coin is flipped $28$ times, either $15$ heads will have occurred or $14$ tails will have occurred, but not both. The probability that $k$ heads occurred after these flips is $$\frac{\binom{28}{k}}{2^{28}}$$ The probability $a$ that at least $15$ heads have occurred is $$a= b – \frac{\binom{28}{14}}{2^{28}}$$ where $b$ is the probability that at least $14$ heads have occurred. $b$ is also equal to the probability that at least $14$ tails have occurred (by symmetry), which is $1 – a$. Solving the resulting system of equations, we find $$a= \frac{2^{25} – 3^3 \cdot 5^2 \cdot 17 \cdot 19 \cdot 23}{2^{26}} = \frac{\cdots 857}{ \cdots 864}$$ Then, the sum of the numerator and denominator of $a \pmod{1000}$ is $\fbox{721}$.

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