Question

Each day John performs the following experiment. He flips a fair coinrepeatedly until he sees a T and counts the number of coin flips needed.(a) Approximate the probability that in a year there are at least 3 days when heneeded more than 10 coin flips. Argue why this approximation is appropriate.(b) Approximate the probability that in a year there are more than 50 days whenhe needed exactly 3 coin flips. Argue why this approximation is appropriate.

Answer 1

Answer:

a. 0.002942

b. 0.244335

Step-by-step explanation:

Suppose the coin is a fair coin

a. Probability that in a year, there are at least 3 days when he needed more than 10 coin flips

First we need to find

$P = P(x>10) = 1-P(x\leq 10)\\=1-[P(x=1)+P(x=2)+...+P(x=10)]\\=1-(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}} )\\=1-\frac{1}{2}(1+\frac{1}{2}+...+\frac{1}{2^9})\\=1-[\frac{1}{2}*\frac{1-(\frac{1}{2})^{10}}{1-\frac{1}{2}}]\\=1-[1-\frac{1}{2^{10}}]=\frac{1}{2^{10}}=\frac{1}{1024}$

For one year, we assume 365 days, therefore

$n=365, p=\frac{1}{1024}\\\alpha _x = 365*\frac{1}{1024} =\frac{365}{1024}=0.356445\\\beta _x=\sqrt{np(1-p)}=\sqrt{365*\frac{1}{1024}*\frac{365}{1024}} =0.596739$

Here, we want to find

$P(x\geq 3)= 1-P(x\leq 2)$

$P(x\leq 2)="=NORMDIST(2, 0.356445, 0.596739,1)"\\=0.997058\\P(x\geq 3)=1-0.997058=0.002942$

b. Probability that in a years time, there are more than 50 days when he needed exactly 3 coin flips

$P = P(x=3)=\frac{1}{2^3}=\frac{1}{8}\\n=365\\\alpha_x=nP=\frac{365}{8}=45.625 \\\beta_x=\sqrt{nP(1-P)} =\sqrt{\frac{365}{8}*\frac{7}{8}} =6.318376\\P(x>50)=1-P(x\leq 50)\\P(x\leq 50)="=NORMDIST(50,45.625,6.318376,1)"=0.755665$

Therefore

$P(x>50)= 1-0.755665=0.244335$