- #1
mathmari
Gold Member
MHB
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Hey! ![Smiling face with smiling eyes :blush: 😊](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f60a.png)
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:
![Smiling face with smiling eyes :blush: 😊](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f60a.png)
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: