In Problem 5, for n = 3, if the coin is assumed fair, what are the probabilities associated with the values that X can take on?

In Problem

SnakeOil

Answered by Aaron 2 years ago

From problem 5, if n = 3, then we have 3 coin flips and possible values of X = {3,1,-1,-3}. To find the different probabilities we use the binomial theorem,

let *T* be the number of tails, and *n* be the number of coin flips then

$$C(n,T)p^T(1-p)^{n-T}$$

where p is 1/2 since it is a fair coin so for the situation in which we have 3 heads and no tails,

$$P(X=3)=C(3,0)\bigg(\frac{1}{2}\bigg)^0 \bigg(1-\frac{1}{2}\bigg)^{3-0}=\frac{1}{8}$$

for 2 heads and 1 tail,

$$P(X=1)=C(3,1)\bigg(\frac{1}{2}\bigg)^1 \bigg(1-\frac{1}{2}\bigg)^{3-1}=\frac{3}{8}$$

for 1 head and 2 tails,

$$P(X=-1)=C(3,2)\bigg(\frac{1}{2}\bigg)^2 \bigg(1-\frac{1}{2}\bigg)^{3-2}=\frac{3}{8}$$

for 0 heads and 3 tails,

$$P(X=-3)=C(3,3)\bigg(\frac{1}{2}\bigg)^3 \bigg(1-\frac{1}{2}\bigg)^{3-3}=\frac{1}{8}$$

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Asked: 2 years ago