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## Sunday, 21 August 2011

### Math GRE - #20

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

• $\dfrac{1}{4}$
• $\dfrac{1}{3}$
• $\dfrac{87}{256}$
• $\dfrac{23}{64}$
• $\dfrac{93}{256}$

Solution :

The answer is $\dfrac{93}{256}$.

The chance of exactly 4 heads and 4 tails is: $\left(\frac{1}{2}\right)^8\cdot\binom{8}{4}=\frac{70}{256}.$

Thus, the chance of having more tails than heads or having more heads than tails is: $1-\frac{70}{256}=\frac{186}{256}.$

Because of symmetry, the probability of having more heads than tails is: $\frac{1}{2}\frac{186}{256}=\frac{93}{256}.$

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