The random variable [tex]X[/tex] has the following probability mass function:
[tex]p_X(x)=\begin{cases}0.63&\text{for }x=1\\0.37&\text{for }x=-1\\0&\text{otherwise}\end{cases}[/tex]
The expected value of [tex]X[/tex] is then
[tex]\mathbb E[X]=\displaystyle\sum_xxp(x)=p(1)-p(-1)=0.26[/tex]
Answer:
[tex] E(X) = 1*0.63 -1*0.37 = 0.26[/tex]
So we expect to win around 0.26 for each game that we play on this game.
Step-by-step 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:
[tex] E(X) =\sum_{i=1}^n X_i P(X_i) [/tex]
For this case we have the following probabilities:
[tex] P(X=1) = 0.63, P(X=-1) = 0.37[/tex]
And then we can replace like this:
[tex] E(X) = 1*0.63 -1*0.37 = 0.26[/tex]
So we expect to win around 0.26 for each game that we play on this game.
