a) Remember that when you bet $1 and lose the bet,you lose the $1. Therefore, you have to earn $2 just to walk out with a net gain of $1. So the answer is:

\(\displaystyle{P}{\left\lbrace{X}{>}{0}\right\rbrace}={\left({\frac{{{18}}}{{{38}}}}\right)}+{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\right)}={.5918}\)

b) To get this answer, you really need to do part first. I'll come back to this after that.

c) Remember that there is a chance that you can lose some dollars when playing. So the answer is as follows:

\(\displaystyle{E}{\left[{X}\right]}=-{3}{\left[{P}{\left(-{3}\right)}\right]}+-{2}{\left[{P}{\left(-{2}\right)}\right]}+-{1}{\left[{P}{\left(-{1}\right)}\right]}+{0}{\left[{P}{\left({0}\right)}\right]}+{1}{\left[{P}{\left({1}\right)}\right]}\)

\(\displaystyle=-{3}{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\right)}+{\left(-{2}\right)}{\left({0}\right)}+{\left(-{1}\right)}{\left[{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\right)}+{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\right)}\right]}+{0}{\left({0}\right)}+{1}{\left[{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\right)}+{\left({\frac{{{18}}}{{{38}}}}\right)}\right]}\)

\(\displaystyle=-{3}{\left({.1458}\right)}+{0}+{\left(-{1}\right)}{\left({.2624}\right)}+{0}+{1}{\left({.5918}\right)}\)

\(\displaystyle=-{.108}\)

Now, going back to part b. This strategy is NOT a"winning strategy" because the expected winnings is a negative value, hence you really have an expected loss.

\(\displaystyle{P}{\left\lbrace{X}{>}{0}\right\rbrace}={\left({\frac{{{18}}}{{{38}}}}\right)}+{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\right)}={.5918}\)

b) To get this answer, you really need to do part first. I'll come back to this after that.

c) Remember that there is a chance that you can lose some dollars when playing. So the answer is as follows:

\(\displaystyle{E}{\left[{X}\right]}=-{3}{\left[{P}{\left(-{3}\right)}\right]}+-{2}{\left[{P}{\left(-{2}\right)}\right]}+-{1}{\left[{P}{\left(-{1}\right)}\right]}+{0}{\left[{P}{\left({0}\right)}\right]}+{1}{\left[{P}{\left({1}\right)}\right]}\)

\(\displaystyle=-{3}{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\right)}+{\left(-{2}\right)}{\left({0}\right)}+{\left(-{1}\right)}{\left[{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\right)}+{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\right)}\right]}+{0}{\left({0}\right)}+{1}{\left[{\left({\left({\frac{{{20}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\cdot{\left({\frac{{{18}}}{{{38}}}}\right)}\right)}+{\left({\frac{{{18}}}{{{38}}}}\right)}\right]}\)

\(\displaystyle=-{3}{\left({.1458}\right)}+{0}+{\left(-{1}\right)}{\left({.2624}\right)}+{0}+{1}{\left({.5918}\right)}\)

\(\displaystyle=-{.108}\)

Now, going back to part b. This strategy is NOT a"winning strategy" because the expected winnings is a negative value, hence you really have an expected loss.