Answer:
a.[tex]\int(x_i)=p(X=x_i)[/tex]
b.[tex]P(X>0)=0.5918[/tex]
C.No
Explanation:
Let X be the probability of a net gain.
a.Given that [tex]X[/tex] is a discrete random variable,[tex]\int(x_i)[/tex] is its probability mass function defined as:
[tex]\int(x_i)=p(X=x_i)[/tex] #for all [tex]i[/tex].
b. Possible outcomes of X are [tex]1,-1,-3[/tex]. A win of $1 occurs when the gambler wins/loses the first bet but is successful in the next two. A win of -$1 occurs in the, if the gambler loses the first and second but wins [tex]3^r^d[/tex] or loses first and third but wins [tex]2^n^d[/tex]. The gambler wins -$3 if he loses all three bets.
[tex]P(X>0)=P(X=1)=\frac{18}{38}+\frac{20}{38}(\frac{18}{38})^2=\frac{4059}{6859}\\=0.5918[/tex]
c.No. The gambler has a more than 50% probability of winning. The winnings are however high risk and less than [tex]zero.[/tex]
