Question 1

A company bids on two contracts.It anticipates a profit of $50,000 if it gets the larger contract and a profit of $20,000 if it gets the smaller contract.It estimates that there's a 20% chance of winning the larger contract and a 60% chance of winning the smaller contract. Create a probability model for the company's profit.Assume that the contracts will be awarded independently.
A) $$\begin{array} { l | l c c c } 
\text { Profit } & \$ 0 & \$ 20,000 & \$ 50,000 & \$ 70,000 \
\hline \text { P(Profit) } & 0.32 & 0.48 & 0.08 & 0.12
\end{array}$$
B) $$\begin{array} { l | l c c c } 
\text { Profit } & \$ 0 & \$ 20,000 & \$ 50,000 & \$ 70,000 \
\hline \text { P(Profit) } & 0.08 & 0.6 & 0.2 & 0.12
\end{array}$$
C) $$\begin{array} { l | c c c } 
\text { Profit } & \$ 0 & \$ 20,000 & \$ 50,000 \
\hline \text { P(Profit) } & 0.2 & 0.6 & 0.2
\end{array}$$
D) $$\begin{array} { l | l c c c } 
\text { Profit } & \$ 0 & \$ 20,000 & \$ 50,000 & \$ 70,000 \
\hline \text { P(Profit) } & 0.32 & 0.48 & 0.08 & 0.8
\end{array}$$
E) $$\begin{array} { l | c c c } 
\text { Profit } & \$ 0 & \$ 20,000 & \$ 50,000 \
\hline \text { P(Profit) } & 0.32 & 0.48 & 0.08
\end{array}$$

Accepted Answer

The probabilities are calculated as follows: - P($0) = P(not winning the larger contract) * P(not winning the smaller contract) = 0.8 * 0.4 = 0.32.- P($20,000) = P(not winning the larger contract) * P(winning the smaller contract) = 0.8 * 0.6 = 0.48.- P($50,000) = P(winning the larger contract) * P(not winning the smaller contract) = 0.2 * 0.4 = 0.08.- P($70,000) = P(winning the larger contract) * P(winning the smaller contract) = 0.2 * 0.6 = 0.12.

Question 2

$$\begin{array} { c | c c c } 
x & 20 & 40 & 60 \
\hline P ( X = x ) & 0.25 & 0.30 & 0.45
\end{array}$$
A)44
B)40
C)50
D)60
E)55

Accepted Answer

Mean = 20 * 0.25 + 40 * 0.30 + 60 * 0.45 = 44

Question 3

$$\begin{array} { r | l c r r } 
\mathrm { x } & 4 & 8 & 12 & 16 \
\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 0.1 & 0.4 & 0.1 & 0.4
\end{array}$$
A)20.00
B)10.00
C)11.20
D)0.25
E)9.80

Accepted Answer

The expected value (mean) of a discrete random variable is calculated using the formula $E(X) = \sum [x \cdot P(X=x)]$. For the given table, this calculation is $E(X) = (4 \cdot 0.1) + (8 \cdot 0.4) + (12 \cdot 0.1) + (16 \cdot 0.4) = 0.4 + 3.2 + 1.2 + 6.4 = 11.2$.

Question 4

A carnival game offers a(n)$80 cash prize for anyone who can break a balloon by throwing a dart at it.It costs $5 to play and you're willing to spend up to $20 trying to win.You estimate that you have a(n)10% chance of hitting the balloon on any throw. Create a probability model for the number of darts you will throw.Assume that throws are independent of each other.Round to four decimal places if necessary.
A) $$\begin{array} { l | c r c c } 
\text { Number of Darts } & 1 & 2 & 3 & 4 \
\hline \text { P(Number of Darts) } & 0.1 & 0.09 & 0.0810 & 0.0656
\end{array}$$
B) $$\begin{array} { l | c c c } 
\text { Number of Darts } & 1 & 2 & 3 \
\hline \text { P(Number of Darts) } & 0.1 & 0.09 & 0.0810
\end{array}$$
C) $$\begin{array} { l | c c c c c } 
\text { Number of Darts } & 1 & 2 & 3 & 4 & 5 \
\hline \mathrm { P } \text { (Number of Darts) } & 0.1 & 0.09 & 0.0810 & 0.7290 & 0.0656
\end{array}$$
D) $$\begin{array} { l | c r c c } 
\text { Number of Darts } & 1 & 2 & 3 & 4 \
\hline \text { P(Number of Darts) } & 0.1 & 0.09 & 0.0810 & 0.7290
\end{array}$$
E) $$\begin{array} { l | c r c c } 
\text { Number of Darts } & 1 & 2 & 3 & 4 \
\hline \mathrm { P } ( \text { Number of Darts } ) & 0.1 & 0.1 & 0.1 & 0.7
\end{array}$$

Accepted Answer

The answer of A carnival game offers a(n)$80 cash prize...

Question 5

A company is interviewing applicants for managerial positions.They plan to hire two people.They have already rejected most candidates and are left with a group of 9 applicants of whom 4 are women.Unable to differentiate further between the applicants,they choose two people at random from this group of 9.Let the random variable X be the number of men that are chosen.Find the probability model for X.
A) $$\begin{array} { l | c c c } 
\text { Number men } & 0 & 1 & 2 \
\hline \text { P(Number men) } & 0.167 & 0.410 & 0.278
\end{array}$$
B) $$\begin{array} { l | c c c } 
\text { Number men } & 0 & 1 & 2 \
\hline \text { P(Number men) } & 0.198 & 0.494 & 0.309
\end{array}$$
C) $$\begin{array} { l | c c c } 
\text { Number men } & 0 & 1 & 2 \
\hline \text { P(Number men) } & 0.278 & 0.556 & 0.167
\end{array}$$
D) $$\begin{array} { l | c c c } 
\text { Number men } & 0 & 1 & 2 \
\hline \text { P(Number men) } & 0.167 & 0.278 & 0.278
\end{array}$$
E) $$\begin{array} { c | c c c } 
\text { Number men } & 0 & 1 & 2 \
\hline \text { P(Number men) } & 11.077 & 66.462 & 66.462
\end{array}$$

Accepted Answer

The probability of choosing 0 men is 4/9 * 5/8 = 20/72 = 5/18. The probability of choosing 1 man is 4/9 * 4/8 = 16/72 = 4/18. The probability of choosing 2 men is 5/9 * 4/8 = 20/72 = 5/18.

Question 6

Hugh buys $8000 worth of stock in an electronics company which he hopes to sell afterward at a profit.The company is developing a new laptop computer and a new desktop computer.If it releases both computers before its competitor,the value of Hugh's stock will jump to $21,000.If it releases one of the computers before its competitor,the value of Hugh's stock will jump to $17,000.If it fails to release either computer before its competitor,Hugh's stock will be worth only $5000.Hugh believes that there is a 40% chance that the company will release the laptop before its competitor and a 50% chance that the company will release the desktop before its competitor. Create a probability model for Hugh's profit.Assume that the development of the laptop and the development of the desktop are independent events.
A) $$\begin{array} { l | c c c } 
\text { Profit } & \$ 21,000 & \$ 17,000 & \$ 5000 \
\hline \text { P(Profit } ) & 0.2 & 0.5 & 0.3
\end{array}$$
B) $$\begin{array} { | l | c c c | } 
\hline \text { Profit } & \$ 13,000 & \$ 9000 & - \$ 3000 \
\hline \text { P(Profit } ) & 0.2 & 0.5 & 0.3
\end{array}$$
C) $$\begin{array} { | l | c c c | } 
\hline \text { Profit } & \$ 13,000 & \$ 9000 & - \$ 3000 \
\hline \text { P(Profit } ) & 0.2 & 0.2 & 0.3
\end{array}$$
D) $$\begin{array} { l | c c c } 
\text { Profit } & \$ 21,000 & \$ 17,000 & \$ 5000 \
\hline \text { P(Profit) } & 0.2 & 0.9 & 0.3
\end{array}$$
E) $$\begin{array} {            l | c c c } 
\text { Profit } & \$ 13,000 & \$ 9000 & - \$ 3000 \
\hline \text { P(Profit } ) & 0.9 & 0.2 & 0.3
\end{array}$$

Accepted Answer

The profits are calculated by subtracting the initial investment ($8000) from the potential values of the stock ($21,000, $17,000, and $5,000), resulting in profits of $13,000, $9,000, and -$3,000, respectively. The probability of both products being released first (and thus earning the highest profit) is the product of their independent probabilities (0.4 for the laptop and 0.5 for the desktop), which is 0.4 * 0.5 = 0.2. The probability of neither being released first is the product of their failure probabilities, 0.6 * 0.5 = 0.3. The remaining probability (1 - 0.2 - 0.3 = 0.5) corresponds to the scenario where only one product is released first.

Question 7

You pick a card from a deck.If you get a face card,you win $15.If you get an ace,you win $20 plus an extra $60 for the ace of hearts.For any other card you win nothing. Create a probability model for the amount you win at this game.
A) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 0 & \$ 15 & \$ 20 & \$ 60 \
\hline \mathrm { P } \text { (Amount won) } & \frac { 39 } { 52 } & \frac { 4 } { 52 } & \frac { 4 } { 52 } & \frac { 1 } { 52 }
\end{array}$$
B) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 0 & \$ 15 & \$ 20 & \$ 80 \
\hline \mathrm { P } \text { (Amount won) } & \frac { 36 } { 52 } & \frac { 12 } { 52 } & \frac { 3 } { 52 } & \frac { 1 } { 52 }
\end{array}$$
C) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 0 & \$ 15 & \$ 20 & \$ 80 \
\hline \mathrm { P } \text { (Amount won) } & \frac { 32 } { 52 } & \frac { 16 } { 52 } & \frac { 3 } { 52 } & \frac { 1 } { 52 }
\end{array}$$
D) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 0 & \$ 15 & \$ 20 & \$ 60 \
\hline \text { P(Amount won) } & \frac { 36 } { 52 } & \frac { 12 } { 52 } & \frac { 3 } { 52 } & \frac { 1 } { 52 }
\end{array}$$
E) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 0 & \$ 15 & \$ 20 & \$ 80 \
\hline \text { P(Amount won) } & \frac { 36 } { 52 } & \frac { 12 } { 52 } & \frac { 4 } { 52 } & \frac { 1 } { 52 }
\end{array}$$

Accepted Answer

The answer of You pick a card from a deck.If...

Question 8

In a box of 7 batteries,3 are dead.You choose two batteries at random from the box.Let the random variable X be the number of good batteries you get.Find the probability model for X.
A) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.184 & 0.490 & 0.327
\end{array}$$
B) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.143 & 0.571 & 0.286
\end{array}$$
C) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.143 & 0.286 & 0.286
\end{array}$$
D) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.286 & 0.571 & 0.286
\end{array}$$
E) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.067 & 0.467 & 0.467
\end{array}$$

Accepted Answer

When choosing two batteries at random, there are a total of (7 choose 2) = 21 possible outcomes. 
We can find the probability of each possible value of X as follows:

- X = 0: There is only 1 way to choose 2 dead batteries out of 3, so P(X=0) = (3 choose 2)/(7 choose 2) = 3/21.
- X = 1: There are 2 ways to choose 1 good battery and 1 dead battery (either pick the two good ones or pick one good and one dead), so P(X=1) = 2*(4 choose 1)*(3 choose 1)/(7 choose 2) = 24/21.
- X = 2: There is only 1 way to choose 2 good batteries out of 4, so P(X=2) = (4 choose 2)/(7 choose 2) = 6/21.

Thus, the probability model for X is:
P(X=0) = 3/21
P(X=1) = 24/21
P(X=2) = 6/21

Note that the probabilities add up to 1, as they should for any probability model.

Question 9

A couple plans to have children until they get a boy,but they agree that they will not have more than four children even if all are girls. Create a probability model for the number of children they will have.Assume that boys and girls are equally likely.
A) $$\begin{array} { l | c c c c } 
\text { Children } & 1 & 2 & 3 & 4 \
\hline \text { P(Children) } & 0.5 & 0.25 & 0.125 & 0.125
\end{array}$$
B) $$\begin{array} { l | c c c c c } 
\text { Children } & 1 & 2 & 3 & 4 & 5 \
\hline \text { P(Children) } & 0.5 & 0.25 & 0.125 & 0.0625 & 0.0625
\end{array}$$
C) $$\begin{array} { l | c c c } 
\text { Children } & 1 & 2 & 3 \
\hline \mathrm { P } \text { (Children) } & 0.5 & 0.25 & 0.25
\end{array}$$
D) $$\begin{array} { l | c c c c } 
\text { Children } & 1 & 2 & 3 & 4 \
\hline \mathrm { P } \text { (Children) } & 0.5 & 0.25 & 0.125 & 0.0625
\end{array}$$
E) $$\begin{array} { l | c c c c } 
\text { Children } & 1 & 2 & 3 & 4 \
\hline \text { P(Children) } & 0.25 & 0.25 & 0.25 & 0.25
\end{array}$$

Accepted Answer

The probability of having a boy (and thus stopping) is 0.5 for the first child. If the first is a girl, the probability of having a boy as the second child is 0.5, but we must also account for the probability of the first being a girl (0.5), making it 0.5 * 0.5 = 0.25 for the second child. This pattern continues, with the probability halving each time. However, if they have three girls in a row (probability 0.125), they will stop at four children regardless, making the probability of having four children 0.125, not halving again because they stop regardless of the fourth child's gender.

Question 10

You roll a fair die.If you get a number greater than 4,you win $70.If not,you get to roll again.If you get a number greater than 4 the second time,you win $30.Otherwise you win nothing. Create a probability model for the amount you win at this game.
A) $$\begin{array} { l | c c c } 
\text { Amount won } & \$ 70 & \$ 30 & \$ 0 \
\hline \text { P(Amount won) } & \frac { 2 } { 6 } & \frac { 8 } { 36 } & \frac { 16 } { 36 }
\end{array}$$
B) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 100 & \$ 70 & \$ 30 & \$ 0 \
\hline \mathrm { P } \text { (Amount won) } & \frac { 4 } { 36 } & \frac { 8 } { 36 } & \frac { 8 } { 36 } & \frac { 16 } { 36 }
\end{array}$$
C) $$\begin{array} { l | c c c } 
\text { Amount won } & \$ 70 & \$ 30 & \$ 0 \
\hline \mathrm { P } \text { (Amount won) } & \frac { 2 } { 6 } & \frac { 2 } { 6 } & \frac { 2 } { 6 }
\end{array}$$
D) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 100 & \$ 70 & \$ 30 & \$ 0 \
\hline \mathrm { P } \text { (Amount won) } & \frac { 4 } { 36 } & \frac { 2 } { 36 } & \frac { 2 } { 36 } & \frac { 16 } { 36 }
\end{array}$$
E) $$\begin{array} { l | c c } 
\text { Amount won } & \$ 70 & \$ 30 \
\hline \text { P(Amount won) } & \frac { 2 } { 6 } & \frac { 4 } { 6 }
\end{array}$$

Accepted Answer

The probability of winning $70 is $\frac{2}{6}$ because there are 2 outcomes (5 or 6) out of 6 that result in winning $70 on the first roll. The probability of winning $30 is $\frac{4}{6} \times \frac{2}{6} = \frac{8}{36}$ because you must first not win on the first roll (4 outcomes out of 6) and then win on the second roll (2 outcomes out of 6). The probability of winning $0 is $\frac{4}{6} \times \frac{4}{6} = \frac{16}{36}$ because you must not win on both rolls, each having 4 outcomes out of 6 that don't result in a win.

Question 11

An insurance policy costs $400,and will pay policyholders $10,000 if they suffer a major injury (resulting in hospitalization),or $2,000 if they suffer a minor injury (resulting in lost time from work).The company estimates that each year 1 in every 3,000 policyholders may have a major injury,and 1 in 1,000 a minor injury. Create a probability model for the company's profit on this policy.
A) $$\begin{array} { l | c | c | c } 
\text { Profit } & \$ 400 & - \$ 10,400 & - \$ 2,400 \
\hline \text { P(profit) } & 0.9987 & 0.0003 & 0.001
\end{array}$$
B) $$\begin{array} { l | c | c | c } 
\text { Profit } & \$ 400 & \$ 10,000 & \$ 2,000 \
\hline \text { P(profit) } & 0.9987 & 0.0003 & 0.001
\end{array}$$
C) $$\begin{array} { l | c | c | c } 
\text { Profit } & \$ 400 & \$ 10,400 & \$ 2,400 \
\hline \text { P(profit) } & 0.9987 & 0.0003 & 0.001
\end{array}$$
D) $$\begin{array} { l | c | c | c } 
\text { Profit } & \$ 400 & \$ 9,600 & \$ 1,600 \
\hline \text { P(profit) } & 0.9987 & 0.0003 & 0.001
\end{array}$$
E) $$\begin{array} { l | c | c | c } 
\text { Profit } & \$ 400 & - \$ 9,600 & - \$ 1,600 \
\hline \text { P(profit) } & 0.9987 & 0.0003 & 0.001
\end{array}$$

Accepted Answer

The profit is calculated as the premium minus the payout. For no injury, the profit is $400. For a major injury, the profit is $400 - $10,000 = -$9,600. For a minor injury, the profit is $400 - $2,000 = -$1,600. The probabilities are given as 0.9987 for no injury, 0.0003 for a major injury, and 0.001 for a minor injury.

Question 12

Your school's soccer team plays two games against another soccer team.The probability that your team wins the first game is 0.6.If your team wins the first game,the probability that they also win the second game is 0.7.If your team loses the first game,the probability that they win the second game is 0.5.Let the random variable X be the number of games won by your team.Find the probability model for X.
A) $$\begin{array} { l | c c c } 
\text { Games won } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Games won) } & 0.12 & 0.46 & 0.42
\end{array}$$
B) $$\begin{array} { l | l c c } 
\text { Games won } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Games won) } & 0.2 & 0.2 & 0.42
\end{array}$$
C) $$\begin{array} { l | l c c } 
\text { Games won } & 0 & 1 & 2 \
\hline \text { P(Games won) } & 0.2 & 0.5 & 0.3
\end{array}$$
D) $$\begin{array} { l | l c c } 
\text { Games won } & 0 & 1 & 2 \
\hline \text { P(Games won) } & 0.2 & 0.18 & 0.42
\end{array}$$
E) $$\begin{array} { l | l c c } 
\text { Games won } & 0 & 1 & 2 \
\hline \text { P(Games won) } & 0.2 & 0.38 & 0.42
\end{array}$$

Accepted Answer

The probability of winning 0 games is 0.4 * 0.5 = 0.2. The probability of winning 1 game is (0.6 * 0.3) + (0.4 * 0.5) = 0.18 + 0.2 = 0.38. The probability of winning 2 games is 0.6 * 0.7 = 0.42.

Question 13

Consider a game that consists of dealing out a hand of two random cards from a deck of four cards.The deck contains the Ace of Spades (As),the Ace of Hearts (Ah),the King of Spades (Ks)and the 9 of Hearts (9h).Aces count as 1 or 11.Kings count as 10.You are interested in the total count of the two cards,with a maximum count of 21 (that is,AsAh = 12).Let X be the sum of the two cards.Find the probability model for X.
A) $$\begin{array} { l | l c r r } 
\mathrm { X } & 1 & 9 & 10 & 11 \
\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 4 & 1 / 4 & 1 / 4 & 1 / 4
\end{array}$$
B) $$\begin{array} { l | l r r r } 
\mathrm { X } & 12 & 19 & 20 & 21 \
\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 6 & 1 / 6 & 1 / 3 & 1 / 3
\end{array}$$
C) $$\begin{array} { l | l r r r r } 
\mathrm { X } & 12 & 19 & 20 & 21 & 22 \
\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 5 & 1 / 5 & 1 / 5 & 1 / 5 & 1 / 5
\end{array}$$
D) $$\begin{array} { l | l r r r r } 
\mathrm { X } & 2 & 12 & 19 & 20 & 21 \
\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 12 & 1 / 12 & 1 / 6 & 1 / 3 & 1 / 3
\end{array}$$
E) $$\begin{array} { l | l r r r } 
\mathrm { X } & 12 & 19 & 20 & 21 \
\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 1 / 4 & 1 / 4 & 1 / 4 & 1 / 4
\end{array}$$

Accepted Answer

Since there are only four cards, we can list all the possible outcomes and their corresponding sum: 
- AsAh = 12 
- AsKs = 12 
- AhKs = 11 
- As9h = 10 
- Ah9h = 10 
- Ks9h = 19

Therefore, the probability model for X is: 
- P(X=12) = 2/6 = 1/3 (two possible outcomes: AsAh or AsKs) 
- P(X=11) = 1/6 (one possible outcome: AhKs) 
- P(X=10) = 2/6 = 1/3 (two possible outcomes: As9h or Ah9h) 
- P(X=19) = 1/6 (one possible outcome: Ks9h)

Note that the sum of all probabilities is 1.

Question 14

You have arranged to go camping for two days in March.You believe that the probability that it will rain on the first day is 0.5.If it rains on the first day,the probability that it also rains on the second day is 0.5.If it doesn't rain on the first day,the probability that it rains on the second day is 0.3.Let the random variable X be the number of rainy days during your camping trip.Find the probability model for X.
A) $$\begin{array} { l | c c c } 
\text { Rainy days } & 0 & 1 & 2 \
\hline \text { P(Rainy days) } & 0.35 & 0.5 & 0.15
\end{array}$$
B) $$\begin{array} { l | c c c } 
\text { Rainy days } & 0 & 1 & 2 \
\hline \text { P(Rainy days) } & 0.35 & 0.25 & 0.25
\end{array}$$
C) $$\begin{array} { l | l l c } 
\text { Rainy days } & 0 & 1 & 2 \
\hline \text { P(Rainy days) } & 0.25 & 0.5 & 0.25
\end{array}$$
D) $$\begin{array} { l | c c c } 
\text { Rainy days } & 0 & 1 & 2 \
\hline \text { P(Rainy days) } & 0.35 & 0.15 & 0.25
\end{array}$$
E) $$\begin{array} { l | l l l } 
\text { Rainy days } & 0 & 1 & 2 \
\hline \text { P(Rainy days) } & 0.35 & 0.4 & 0.25
\end{array}$$

Accepted Answer

The answer of You have arranged to go camping for...

Question 15

You roll a pair of fair dice.If you get a sum greater than 10 you win $50.If you get a double you win $40.If you get a double and a sum greater than 10 you win $90.Otherwise you win nothing. Create a probability model for the amount you win at this game.
A) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 0 & \$ 40 & \$ 50 & \$ 90 \
\hline \text { P(Amount won) } & \frac { 28 } { 36 } & \frac { 5 } { 36 } & \frac { 2 } { 36 } & \frac { 1 } { 36 }
\end{array}$$
B) $$\begin{array} { l | l l l } 
\text { Amount won } & \$ 0 & \$ 40 & \$ 50 \
\hline \mathrm { P } \text { (Amount won) } & \frac { 27 } { 36 } & \frac { 6 } { 36 } & \frac { 3 } { 36 }
\end{array}$$
C) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 0 & \$ 40 & \$ 50 & \$ 90 \
\hline \text { P(Amount won) } & \frac { 27 } { 36 } & \frac { 5 } { 36 } & \frac { 3 } { 36 } & \frac { 1 } { 36 }
\end{array}$$
D) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 0 & \$ 40 & \$ 50 & \$ 90 \
\hline \text { P(Amount won) } & \frac { 26 } { 36 } & \frac { 6 } { 36 } & \frac { 3 } { 36 } & \frac { 1 } { 36 }
\end{array}$$
E) $$\begin{array} { l | c c c c } 
\text { Amount won } & \$ 0 & \$ 40 & \$ 50 & \$ 90 \
\hline \text { P(Amount won) } & \frac { 27 } { 36 } & \frac { 6 } { 36 } & \frac { 2 } { 36 } & \frac { 1 } { 36 }
\end{array}$$

Accepted Answer

The probability of getting a sum greater than 10 is 2/36, the probability of getting a double is 5/36, and the probability of getting a double and a sum greater than 10 is 1/36. The probability of winning nothing is 28/36.

Question 16

A carnival game offers a(n)$80 cash prize for anyone who can break a balloon by throwing a dart at it.It costs $5 to play and you're willing to spend up to $20 trying to win.You estimate that you have a(n)12% chance of hitting the balloon on any throw. Create a probability model for the amount you will win.Assume that throws are independent of each other.Round to four decimal places if necessary.
A) $$\begin{array} { l | l c c c c } 
\text { Amount won } & \$ 75 & \$ 70 & \$ 65 & \$ 60 & - \$ 20 \
\hline \mathrm { P } \text { (Amount won } ) & 0.12 & 0.1056 & 0.0929 & 0.0818 & 0.0720
\end{array}$$
B) $$\begin{array} { l | l c c c } 
\text { Amount won } & \$ 75 & \$ 70 & \$ 65 & \$ 60 \
\hline \mathrm { P } \text { (Amount won) } & 0.12 & 0.1056 & 0.0929 & 0.6815
\end{array}$$
C) $$\begin{array} { l | l c c c c } 
\text { Amount won } & \$ 80 & \$ 75 & \$ 70 & \$ 65 & - \$ 20 \
\hline \mathrm { P } \text { (Amount won } ) & 0.12 & 0.1056 & 0.0929 & 0.0818 & 0.0720
\end{array}$$
D) $$\begin{array} { l | l c c c c } 
\text { Amount won } & \$ 75 & \$ 70 & \$ 65 & \$ 60 & - \$ 20 \
\hline \mathrm { P } \text { (Amount won } ) & 0.12 & 0.1056 & 0.0929 & 0.0818 & 0.5997
\end{array}$$
E) $$\begin{array} { l | l c c c c } 
\text { Amount won } & \$ 80 & \$ 75 & \$ 70 & \$ 65 & - \$ 20 \
\hline \mathrm { P } \text { (Amount won } ) & 0.12 & 0.1056 & 0.0929 & 0.0818 & 0.5997
\end{array}$$

Accepted Answer

Let X be the amount you will win. 
The possible outcomes for X are: 
- You win $75 with probability 0.12, since you can win $80 but you spent $5 to play. 
- You lose $5 with probability 0.88, since you don't win anything but you spent $5 to play.

Therefore, the probability model for X is:

|   X   |  75  | -5   |
|-------|------|------|
| P(X)  | 0.12 | 0.88 |

To find the expected value of X, we multiply each outcome by its probability and sum the results:

E(X) = ($75)(0.12) + (-$5)(0.88) = -$3.4

Since the expected value of X is negative, it means that on average you will lose money by playing this game. Therefore, it's not a good choice to spend up to $20 trying to win.

Question 17

$$\begin{array} { l | r r r r } 
\mathrm { x } & 100 & 200 & 300 & 400 \
\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 0.3 & 0.4 & 0.2 & 0.1
\end{array}$$
A)220
B)250
C)210
D)200
E)150

Accepted Answer

The answer of \[\begin{array} { l | r r r...

Question 18

You pick a card from a deck.If you get a club,you win $80.If not,you get to draw again (after replacing the first card).If you get a club the second time,you win $30.Otherwise you win nothing. Create a probability model for the amount you win at this game.
A) $$\begin{array} { l | r r r } 
\text { Amount won } & \$ 80 & \$ 30 & \$ 0 \
\hline \text { P(Amount won) } & \frac { 1 } { 4 } & \frac { 21 } { 64 } & \frac { 27 } { 64 }
\end{array}$$
B) $$\begin{array} { l | c c c c c } 
\text { Amount won } & \$ 80 & \$ 60 & \$ 30 & \$ 0 \
\hline P ( \text { Amount won) } & \frac { 1 } { 4 } & \frac { 9 } { 64 } & \frac { 3 } { 16 } & \frac { 27 } { 64 }
\end{array}$$
C) $$\begin{array} { l | c c c } 
\text { Amount won } & \$ 80 & \$ 30 & \$ 0 \
\hline \mathrm { P } \text { (Amount won) } & \frac { 1 } { 4 } & \frac { 3 } { 16 } & \frac { 9 } { 16 }
\end{array}$$
D) $$\begin{array} { l | c c c c c } 
\text { Amount won } & \$ 110 & \$ 80 & \$ 30 & \$ 0 \
\hline P ( \text { Amount won) } & \frac { 1 } { 16 } & \frac { 3 } { 16 } & \frac { 3 } { 16 } & \frac { 9 } { 16 }
\end{array}$$
E) $$\begin{array} { l | r c c } 
\text { Amount won } & \$ 80 & \$ 30 & \$ 0 \
\hline \text { P(Amount won) } & \frac { 1 } { 4 } & \frac { 1 } { 4 } & \frac { 1 } { 2 }
\end{array}$$

Accepted Answer

The probability model for this game can be constructed using a tree diagram as follows:

- The first level represents the first draw, with a probability of 1/4 for each suit (since there are four suits in a deck).
- If a club is drawn on the first attempt (probability 1/4), the player wins $80.
- If a club is NOT drawn on the first attempt (probability 3/4), the game moves to the second level, representing the second draw. At this level, the probability of drawing a club is again 1/4, since the deck was replaced.
- If a club is drawn on the second attempt (probability 1/4 x 1/4 = 1/16), the player wins an additional $30, for a total of $110.
- If a club is NOT drawn on the second attempt (probability 1 - 1/16 = 15/16), the player wins nothing.

Using this probability model, we can calculate the expected value of the game as follows:

EV = (1/4)($80) + (3/4)(1/4)($30) + (3/4)(15/16)(0)
EV = $20 + $2.81
EV = $22.81

Therefore, the best choice would be C, which represents the expected value of $22.81.

Question 19

$$\begin{array} { c | l l l } 
\mathrm { x } & 0 & 1 & 2 \
\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 0.5 & 0.2 & 0.3
\end{array}$$
A)1.00
B)0.60
C)1.20
D)0.80
E)0.33

Accepted Answer

The answer of \[\begin{array} { c | l l l...

Question 20

$$\begin{array} { r | r } 
\mathrm { x } & \mathrm { P } ( \mathrm { X } = \mathrm { x } ) \
\hline 0 & 0.50 \
1 & 0.12 \
2 & 0.16 \
3 & 0.20 \
4 & 0.02
\end{array}$$
A)1.12
B)0.97
C)1.52
D)1.02
E)1.62

Accepted Answer

The expected value is 0.50*0 + 0.12*1 + 0.16*2 + 0.20*3 + 0.02*4 = 1.12

Quiz 14: Random Variables

$\begin{array} { c | c c c } x & 20 & 40 & 60 \\\hline P ( X = x ) & 0.25 & 0.30 & 0.45\end{array}$

$\begin{array} { r | l c r r } \mathrm { x } & 4 & 8 & 12 & 16 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 0.1 & 0.4 & 0.1 & 0.4\end{array}$

You pick a card from a deck.If you get a face card,you win $15.If you get an ace,you win $20 plus an extra $60 for the ace of hearts.For any other card you win nothing. Create a probability model for the amount you win at this game.

In a box of 7 batteries,3 are dead.You choose two batteries at random from the box.Let the random variable X be the number of good batteries you get.Find the probability model for X.

A couple plans to have children until they get a boy,but they agree that they will not have more than four children even if all are girls. Create a probability model for the number of children they will have.Assume that boys and girls are equally likely.

You roll a fair die.If you get a number greater than 4,you win $70.If not,you get to roll again.If you get a number greater than 4 the second time,you win $30.Otherwise you win nothing. Create a probability model for the amount you win at this game.

You roll a pair of fair dice.If you get a sum greater than 10 you win $50.If you get a double you win $40.If you get a double and a sum greater than 10 you win $90.Otherwise you win nothing. Create a probability model for the amount you win at this game.

$\begin{array} { l | r r r r } \mathrm { x } & 100 & 200 & 300 & 400 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 0.3 & 0.4 & 0.2 & 0.1\end{array}$

You pick a card from a deck.If you get a club,you win $80.If not,you get to draw again (after replacing the first card) .If you get a club the second time,you win $30.Otherwise you win nothing. Create a probability model for the amount you win at this game.

$\begin{array} { c | l l l } \mathrm { x } & 0 & 1 & 2 \\\hline \mathrm { P } ( \mathrm { X } = \mathrm { x } ) & 0.5 & 0.2 & 0.3\end{array}$

$\begin{array} { r | r } \mathrm { x } & \mathrm { P } ( \mathrm { X } = \mathrm { x } ) \\\hline 0 & 0.50 \\1 & 0.12 \\2 & 0.16 \\3 & 0.20 \\4 & 0.02\end{array}$