Question

An unfair coin is flipped. if a head turns​ up, you win​ $1. if a tail turns​ up, you lose​ $1. the probability of a head is . 63 and the probability of a tail is . 37. let x be the random variable for the amount won on a single play of this game. what is the expected value of the​ game?

Answer 1

The random variable
`X` has the following probability mass function:

`p_X(x)=\begin{cases}0.63&\text{for }x=1\\0.37&\text{for }x=-1\\0&\text{otherwise}\end{cases}`

The expected value of
`X` is then

`\mathbb E[X]=\displaystyle\sum_xxp(x)=p(1)-p(-1)=0.26`

Answer 2

`E(X) = 1*0.63 -1*0.37 = 0.26`

So we expect to win around 0.26 for each game that we play on this game.

Explanation:

For this case we can define a random variable who represent the amount of money win or loss X. X=1 if we got a head and X=-1 if we got a tail.

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

And from definition the expected value is defined with this formula:

`E(X) =\sum_(i=1)^n X_i P(X_i)`

For this case we have the following probabilities:

`P(X=1) = 0.63, P(X=-1) = 0.37`

And then we can replace like this:

`E(X) = 1*0.63 -1*0.37 = 0.26`

So we expect to win around 0.26 for each game that we play on this game.