in a game you flip a coin twice, and record the number of heads that occur. you get $15 for 2 heads, pay $3 for 1 head, and pay $2 for no heads. what is the expected value for the amount youll win/lose per turn? write the answer without the $

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Accepted Answer

# Explanation:
## Step1: Calculate the probability of each outcome
When flipping a coin twice, the sample space is $S=\{HH, HT, TH, TT\}$, so $n(S) = 4$.
- Probability of $2$ heads ($P(2H)$): There is $1$ outcome ($HH$) with $2$ heads. So $P(2H)=\frac{1}{4}$.
- Probability of $1$ head ($P(1H)$): There are $2$ outcomes ($HT,TH$) with $1$ head. So $P(1H)=\frac{2}{4}=\frac{1}{2}$.
- Probability of $0$ heads ($P(0H)$): There is $1$ outcome ($TT$) with $0$ heads. So $P(0H)=\frac{1}{4}$.

## Step2: Use the expected - value formula $E(X)=\sum_{i}x_{i}P(x_{i})$
Let $x_{1} = 15$ (for $2$ heads), $x_{2}=- 3$ (for $1$ head, since it's a loss), $x_{3}=-2$ (for $0$ heads, since it's a loss).
$$
\begin{align*}
E(X)&=(15)\times\frac{1}{4}+(-3)\times\frac{1}{2}+(-2)\times\frac{1}{4}\
&=\frac{15}{4}-\frac{3}{2}-\frac{2}{4}\
&=\frac{15 - 6-2}{4}\
&=\frac{15-(6 + 2)}{4}\
&=\frac{7}{4}\
& = 4.25
\end{align*}
$$

# Answer:
$4.25$

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Explanation:

Step1: Calculate the probability of each outcome

Step2: Use the expected - value formula \(E(X)=\sum_{i}x_{i}P(x_{i})\)

Answer: