MHB Game : random variable for net profit

[mathmari]

Hey!

You participate in the following game :
You toss a fair coin until heads falls, but no more than three times. You have to pay $1$ euro for each throw. If your head falls, you win $3$ euros. The random variable $X$ describes your net profit (profit
minus stake). Give the values that $X$ can get and the corresponding probabilities. Calculate $P [X <0]$ and $P [X> 0]$. Make a note of the intermediate steps.

I have done the following :
- If we get Head at the first toss, then X = -1 + 3 = 2 EUR.
- If we get Head at the second toss, then X = -1 -1 + 3 = 1 EUR.
- If we get Head at the third toss, then X = -1 -1 -1 + 3 = 0 EUR.
- If we don't get Head at all at the three first tosses, then X = -1 -1 - 1 = -3 EUR.

So are the possible values for $X$ the $\{-3, 0, 1, 2\}$ ? :unsure:

As for the probabilities do we have the following ?
- $P(X=2) = \frac{1}{2}$ (either head or no head at the first toss)
- $P(X=1) = \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$ (either head or no head at the first toss, either head or no head at the second toss)
- $P(X=0) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ (either head or no head at the first toss, either head or no head at the second toss, either head or no head at the third toss)
- $P(X=0) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ (either head or no head at the first toss, either head or no head at the second toss, either head or no head at the third toss)

Is everything correct and complete? :unsure:

 

[I like Serena]

Hey mathmari!

All correct. (Nod)

I would rephrase the explanations though.
For instance:
- $P(X=2) = \frac{1}{2}$ (head at the first toss)
- $P(X=1) = \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$ (no head at the first toss, and head at the second toss)
(Thinking)

 

[mathmari]

As for the probabilities do we have the following ?
- $P(X=2) = \frac{1}{2}$ (either head or no head at the first toss)
- $P(X=1) = \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$ (either head or no head at the first toss, either head or no head at the second toss)
- $P(X=0) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ (either head or no head at the first toss, either head or no head at the second toss, either head or no head at the third toss)
- $P(X=0) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ (either head or no head at the first toss, either head or no head at the second toss, either head or no head at the third toss)

Ah at the last one Imeant $P(X=-3)$ instead of $P(X=0)$, so we have $P(X=-3) = \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ ( no head at the first toss, and no head at the second toss, and no head at the third toss ), right? :unsure:

To calculate the probabilities $P(X<0)$ and $P(X>0)$ do we do the following ?
\begin{align*}&\{X<0\}=\cup_{{y\in X(\Omega), y<0}}\{X=y\}=\{X=-3\} \\ & \Rightarrow P[X<0]=P[X=-3]=\frac{1}{8} \\ &\{X>0\}=\cup_{{y\in X(\Omega), y>0}}\{X=y\}=\{X=1\} \cup \{X=2\} \\ & \Rightarrow P[X>0]=P[\{X=1\} \cup \{X=2\}]=P[X=1]+P[X=2]=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\end{align*}
Is that correct? Do we have to mention more intermediate steps or are these the desired ones? :unsure:

 

[I like Serena]

Looks correct to me, and I think you have all the desired steps. (Nod)

 

[mathmari]

Looks correct to me, and I think you have all the desired steps. (Nod)

I added some more steps. Is that correct now? :unsure:

 

[I like Serena]

I added some more steps. Is that correct now?

It looks fine to me. (Nod)