Question

You have a biased coin for which p(h)=pp(h)=p. you toss the coin 2020 times. what is the probability that you observe 88 heads and 1212 tails; you observe more than 88 heads and more than 88 tails?

Answer 1

The question describes a binomial probability with p(h) = p, then p(t) = 1 - p and number of trials (n) = 20

The probability of a binomial distribution is given by

$P(x)=\, ^nC_xp^x(1-p)^{n-x}$

Part A:

The probability of observing 8 heads and 12 tails is given by:

$P(8)=\, ^{20}C_8p^8(1-p)^{20-8}=\, ^{20}C_8p^8(1-p)^{12}$



Part B:

You observe more than 8 heads and more than 8 tails, when you observe 9 heads and 11 tails, 10 heads and 10 tails, and 11 heads and 9 tails.

Therefore, the probability of observing more than 8 heads and more than 8 tails is given by:

$P(9)+P(10)+P(11) \\ \\ =\, ^{20}C_9p^9(1-p)^{20-9}+\, ^{20}C_{10}p^{10}(1-p)^{20-10}+\, ^{20}C_{11}p^{11}(1-p)^{20-11} \\ \\ =\, ^{20}C_9p^9(1-p)^{11}+\, ^{20}C_{10}p^{10}(1-p)^{10}+\, ^{20}C_{11}p^{11}(1-p)^{9}$