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A fair coin is to be tossed $8$ times. What is the probability that more of the tosses will result in heads than will result in tails?

$\textbf{Guess:}$ I'm guessing that by symmetry, we can write down the probability $x$ of getting exactly $4$ heads and $4$ tails and then calculate $\dfrac{1}{2}(1-x)$.

So how does one calculate for $x$? I know that it should be a rational number (that is, $\dfrac{?}{2^8}$), but I am not sure how to get the numerator.

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4 Answers 4

up vote 3 down vote accepted

The number in the numerator should be $\displaystyle \left( \begin{array}{c} 8 \\ 4 \end{array} \right) = \frac{8!}{4! ( 8-4)!} = 70$.

Why? Because we have $8$ tosses, and out of these tosses, we have $4$ heads.

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Thank you Andrew! –  user81136 Jun 6 '13 at 1:00

P(getting more Heads in 8 tosses)=(8C5+8C6+8C7+8C8)/(2^8)

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Use Pascal's Triangle to find the numerator. For this particular problem, the numerator is 70.

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Use the binomial distribution to get the probability of getting $k$ heads from $n$ flips:

$$p(n,k) = \binom{n}{k} \left ( \frac12 \right )^k \left ( \frac12 \right )^{n-k} = \binom{n}{k} \left ( \frac12 \right )^n$$

The probability you seek is $p(8,5)+p(8,6)+p(8,7)+p(8,8)$, or

$$\frac{\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}{2^8} = \frac{56+28+8+1}{2^8} = \frac{93}{256}$$

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