Module 4-5 Counting, Probability, And Expected Value Show ✓ Solved

Module 4-5 Counting, Probability, and Expected Value Show

Counting

1. A gift-wrapping store has 8 shapes of boxes, 14 types of wrapping paper, and 12 different bows. How many different gift-wrapping options are available at this store?

2. Four friends go to the movies. How many different ways can they sit in a row?

3. You are creating a 4-letter password using only the 26 lowercase letters from the alphabet. How many unique passwords could you create if:

a. Repeating letters is allowed?

b. Repeating letters are not allowed?

c. What are some passwords that are possible in a. that are not possible in b? What are some passwords that are possible in either scenario?

4. Which of the following do I have to check before using nPr? There may be more than one correct answer. a. b. Repeating is allowed. c. Repeating is not allowed. d. The order matters. e. The order does not matter.

5. Which of the following do I have to check before using nCr? There may be more than one correct answer. a. b. Repeating is allowed. c. Repeating is not allowed. d. The order matters. e. The order does not matter.

6. 7. Two people out of a group of 75 will win tickets to an upcoming concert. How many different groups of two are possible?

8. Barry is hosting a Super Bowl party and offers 7 different kinds of chip dip. If a party goer can choose any number of chip dips for their chips, how many chip dip combinations are possible?

9. There are 30 students in the classroom competing for classroom prizes. Only the first two students whose name is drawn will win a prize. If you are one of the 30 students, what is the probability you will win the top prize?

Probability and Odds

Answer all questions with a fraction in lowest terms. If you’d like, you can also write them as a percent. If you draw one card at random, what is the probability that card is a(n)…

9. Heart?

10. 7 of diamonds?

11. Face card or a club? Given the card is a club, what is the probability a card drawn at random will be a(n)…

12. 8?

13. 10 or ace?

You are choosing two cards, without replacing the first card. What is the probability you choose…

14. A 7 then a 3?

15. Two consecutive fours?

16. Two consecutive diamonds?

You are choosing two cards, replacing the first card in the deck after it has been drawn. What is the probability you choose…

17. A 7 then a 3?

18. Two consecutive fours?

19. Two consecutive diamonds?

20. A card is drawn from a standard deck of 52 cards, and then replaced in the deck. Find the probability that at least one king is drawn by the fourth draw. Round your answer to two decimal places.

21. Bill and Sam are playing a game where they roll a die and draw a card from a deck. What is the probability of rolling a 5 and then drawing a 5 from the deck of cards? Write your answer as a fraction in lowest terms.

22. Suppose on a particular day, the probability of getting into a car accident is 0.04, the probability of being a texter-and-driver is 0.14, and p(car accident or being a texter-and-driver)=0.15. Find the probability a person was in a car accident given that they are a texter-and-driver.

23. Suppose you are playing a game with one die at a casino. You win $1 if you get a 2 or a 6. You lose $2 if you get a 3, win $5 for rolling a 4, and lose $3 for rolling a 1. What is the expected value of this game?

24. Suppose you are playing a game with two dice at a casino. If you roll double ones, you win $1, if you roll double twos, you win $2. This pattern continues, up to winning $6 for double sixes. If you roll anything besides doubles, you lose $1. What is the expected value of the game?

Paper For Above Instructions

In mathematics and statistics, the concepts of counting, probability, and expected value are fundamental tools used to analyze various scenarios. This paper will delve into several problems that encompass these concepts, providing solutions while elaborating on the underlying principles involved in each calculation. By addressing counting problems alongside probability questions, insights into how these aspects interact in real-world situations will be presented.

Counting Problems

1. To find the different gift-wrapping options available in a gift-wrapping store with 8 shapes of boxes, 14 types of wrapping paper, and 12 different bows, the total options can be calculated using the rule of product (multiplication rule).

The total number of different combinations:

Total options = Number of boxes × Number of papers × Number of bows

Total options = 8 × 14 × 12 = 672. Thus, there are 672 different gift-wrapping options available.

2. When determining how many different ways four friends can sit in a row, the problem can be solved using permutations. For four friends, the arrangement can be computed as:

Number of arrangements = 4! = 4 × 3 × 2 × 1 = 24.

This means there are 24 different ways for the four friends to sit in a row.

3. The creation of a 4-letter password using the 26 lowercase letters involves different scenarios based on whether letters can repeat or not:

a. If repeating letters is allowed, the total number of unique passwords is:

Number of passwords = 26^4 = 456976.

b. If repeating letters are not allowed, it is calculated as:

Number of passwords = 26 × 25 × 24 × 23 = 358800.

c. Examples of passwords possible with repeating letters (e.g., 'aaaa', 'abcd') and examples of those not possible when repeating letters are forbidden (e.g., 'abcd' cannot be repeated). If 'abcd' can appear in both scenarios, it illustrates this interaction.

4. For nPr (permutations), checks involve noting:

- Repeating is allowed (No)

- Repeating is not allowed (Yes)

- The order matters (Yes)

- The order does not matter (No)

5. For nCr (combinations), the checks must include:

- Repeating is allowed (No)

- Repeating is not allowed (Yes)

- The order matters (No)

- The order does not matter (Yes)

6. In the problem of choosing 2 winners from 75 people, we use combinations (nCr):

Combinations = 75C2 = 75! / (2!(75-2)!) = 2925.

8. Barry's chip dip combinations arise from choosing any number of dips (including none). This is essentially a power set, concluding in:

Chip dip combinations = 2^7 - 1 = 127.

9. The probability of winning a prize as one of the students can be computed as follows:

- Probability of you winning = 1/30, and your friend winning = 1/29, so the combined probability of both events is:

P(You win and friend wins) = 1/30 × 1/29 = 1/870.

Probability Problems

10. For drawing a heart from a 52-card deck, it equals:

P(Heart) = 13/52 = 1/4.

11. The probability of drawing a 7 of diamonds is:

P(7 of diamonds) = 1/52.

12. To find the probability of a face card or a club, calculate:

- 12 face cards + 13 clubs - 3 face card clubs = 22/52 = 11/26.

13. If a club is drawn, the probability of drawing an 8 is:

P(8 | Club) = 1/13.

14. The probability of drawing a 7 followed by a 3 without replacement is:

- First: 4/52, Second: 4/51, thus:

P(7 then 3) = 4/52 × 4/51 = 1/663.

15. For two consecutive fours, probability is:

- P(4 then 4) = 4/52 × 3/51 = 1/221.

16. For two consecutive diamonds:

- P(Diamond then Diamond) = 13/52 × 12/51.

17. When drawing with replacement, the probability of a 7 then a 3 is:

- P(7 then 3 with replacement) = 4/52 × 4/52.

18. Two consecutive fours with replacement is calculated similarly to previous instances.

19. For drawing two diamonds: P(Diamond | Replacement) = 13/52 x 13/52.

20. The probability of at least one King drawn by the fourth draw can be calculated using the complement rule. Calculate 1 - P(no kings drawn in four draws). The calculations involve probabilities of not drawing a king across four draws and subtracting that from 1.

21. For Bill and Sam rolling a die and drawing a card, the calculations for the combined probability of both events, finding the chance of both rolling a 5 and drawing a 5 yield a fraction in reduced form.

Expected Values

22. Expected value for casino games requires calculating potential winnings against the probabilities of winning and losing scenarios described. The expected value derived will offer insights into whether the game favors the player or the house.

23. The final expected value computation requires weighing the outcomes described, elucidating a net expected profit or loss.

Conclusion

In conclusion, the analyses of counting, probability, and expected value are pivotal in determining outcomes across various scenarios. Through the application of these fundamental concepts, individuals can make more informed decisions based on statistical reasoning. As demonstrated, by systematically addressing each problem, we equip ourselves with practical mathematical skills essential in everyday decision-making.

References

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