You toss a fair coin 10000 times. what are the odds of obtaining more than 5100 tails, approximately?

1 answer:

3 0

This can be solved by using the normal approximation to the binomial distribution.
mean = np = 10.000 * 0.5 = 5,000
The standard deviation is given by:
$S.D.= \sqrt{npq} = \sqrt{5000\times0.5} =50$
$z=\frac{5100-5000}{50}=2$
The probability of obtaining more than 5100 tails is 0.0228 and the probability of obtaining fewer than 5100 tails is 0.9772.
The odds of obtaining more than 5100 tails is therefore:
0.0228:0.9772 = 1:42.86.

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