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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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