Question

section 6.1 homework score: 8.67/21 answered: 9/21 question 12 suppose that you are offered the following \deal.\ you roll a six - sided die. if you roll a 2, 3, 4 or 5, you win $5. otherwise, you pay $10. if you roll a 6, you win $7. a. complete the pdf table. list the x values, where x is the profit, from smallest to largest. round to 4 decimal places where appropriate. probability distribution table x p(x) b. find the expected profit. $ (round to the nearest cent) c. interpret the expected value. if you play many games you will likely win on average very close to $2.83 per game. this is the most likely amount of money you will win. you will win this much if you play a game.

Explanation:

Step1: Calculate probabilities for each outcome

A fair six - sided die has 6 possible outcomes. The probability of rolling a 2, 3, 4, or 5 is $\frac{4}{6}=\frac{2}{3}$, the probability of rolling a 6 is $\frac{1}{6}$, and the probability of rolling a 1 is $\frac{1}{6}$.

Step2: Calculate profit values for each outcome

If rolling a 2, 3, 4, or 5, the profit $X = 5 - 10=- 5$ (since you pay $10$ and win $5$). If rolling a 6, the profit $X = 7 - 10=-3$. If rolling a 1, the profit $X=-10$ (you just pay $10$ and win nothing).

Step3: Create the probability - distribution function (PDF) table

X	P(X)
- 5	$\frac{2}{3}$
- 3	$\frac{1}{6}$

Step4: Calculate the expected value

The formula for the expected value $E(X)=\sum_{i}x_ip_i$. So $E(X)=(-10)\times\frac{1}{6}+(-5)\times\frac{2}{3}+(-3)\times\frac{1}{6}=\frac{-10 - 20 - 3}{6}=\frac{-33}{6}=- 5.5$

Answer:

a.

X	P(X)
- 5	$\frac{2}{3}$
- 3	$\frac{1}{6}$

b. -$5.50$
c. If you play many games, on average, you will lose $5.50$ per game. This means that over the long - run, it is not a favorable game to play as you are expected to lose money.