Answer each of the following problems. Make sure to show your work.

Transcription

1 Answer each of the following problems. Make sure to show your work. 1. A board game requires each player to roll a die. The player with the highest number wins. If a player wants to calculate his or her probability of winning the game, what is the sample space they must use? The sample space is the set of all possible outcomes. Therefore S = [1, 2, 3, 4, 5, 6]. 2. A board game requires each player to draw a card. Each card has a different number on it. The player with the highest number wins. Is the probability of a player winning an independent or dependent event? Why? The event is dependent because the probability that a winning card will be drawn is changed every time a card is drawn. Therefore, the results of the other players draws affect the probability that a particular player will win. 3. A box contains 12 red pens, 14 blue pens, and 6 black pens. If a red pen is drawn first and kept out of the bag, what fraction represents the probability that a black pen will be drawn second? 6/31. There are 6 black pens left and the red pen was removed, so the original sample size of 32 is reduced to A die has been rolled seven times in a row. Each time, the die landed on six. What is the probability that the die will land on six on the eighth roll? Why? The probability of landing on six is 1/6 because there are six sides to the die. The roll is an independent event, so the results of the previous rolls have no effect on the outcome of the eighth roll. 5. The probability that it will rain today is 70%. The probability that it will just be cloudy is 20%, and the probability that it will be sunny and clear is 10%. Is today s weather a biased event? Yes, the event is biased because the outcomes have different probabilities. 6. What is the probability that it will rain or that it will be sunny and clear? These are mutually exclusive events, so we simply add the probabilities together. P= =.8 = 80%.

2 7. If it rains, a gardener has a 40% chance of having a day off on the following day. If it doesn t rain, he has to work the next day. What is the probability that the gardener will have to work tomorrow? P(day off tomorrow) = P(rain)(1-P(day off)) + P(no rain) = (.7)(.6) +.3 = =.72 = 72%. 8. What is the probability that each spinner will land on purple? P(both on purple) = P(first spinner on purple) x P(second spinner on purple) = 2/8 x 2/8 = 4/64 = 1/16.

3 Use the following tree to answer questions What is the value of X? The value of X is 1-.6 = What is the probability that Event D will be the outcome? The probability of event D is equal to.8 x.1 = If A represents that it will snow and there will be a snow day, B represents that it it will snow and there won t be a snow day, C represents that it won t snow and there won t be a snow day, and D represents that it won t snow and there will be a snow day, what is the probability that there won t be a snow day? P(no snow day) = P(B AND C) = (.2)(.6) + (.8)(.9) = =.84 = 84%. 12. What is the probability that there will be a snow day? P(snow day) = P(A AND C) = (.2)(.4) + (.8)(.1) = =.16 = 16% (which is also 1 P(no snow day))

4 13. What probability model should be used when examining simple events? A systematic list can be used to model simple events effectively. 14. A game involves drawing 16 red coins, 12 blue coins and 3 green coins from a bag. A player wins if they draw a blue or green coin and loses if they draw a red coin. Is this game fair? Explain your answer. The game is not fair because the probability of winning is P(blue coin) + P(green coin) = 15/31, whereas the probability of losing is P(red coin) = 16/31. Because it is more likely that a player will lose the game, the game cannot be considered fair. 15. A game involves drawing 16 red coins, 13 blue coins and 3 green coins from a bag. A player wins if they draw a blue or green coin and loses if they draw a red coin. Is this game fair? Explain your answer. The game is fair because the probability of winning is P(blue coin) + P(green coin) = 16/32 = 1/2, whereas the probability of losing is P(red coin) = 16/32 = 1/2. Because the probability of losing the game is the same as the probability of winning the game, the game is considered fair. 16. What model should be used when performing complex probability calculations with multiple AND functions? Explain your answer. A tree diagram should be used. The tree diagram provides easy visual cues for and calculations. All one has to do is follow a path from the root of the tree to a node, multiplying the probabilities on the branches along the way. 17. At a given company, an employee is 42% likely to be male and 58% likely to be a female. An employee is also 30% likely to be an engineer, 50% likely to be a manager, 25% likely to be a researcher and 10% likely to be a technical writer. If Event A = male manager or male engineer, what is event A? Event A = male researcher or male tech writer or female. These are all the possible outcomes that aren t a male manager or a male engineer, and therefore form the complement of A. 18. At a given company, an employee is 42% likely to be male and 58% likely to be a female. An employee is also 30% likely to be an engineer, 50% likely to be a manager, 25% likely to be a researcher and 10% likely to be a technical writer. What is the probability of the union of being an engineer or a technical writer? P(E U TW) = P(E) + P(TW) P(E)P(TW) = (.3)(.1) = =.37 = 37%

5 19. Given Event A, Event B, and Event C. Event A and Event B are mutually exclusive. Event A and Event C are not mutually exclusive. P(A) = 0.45 P(B) = 0.35 P(C) = 0.25 What is the probability of the complement of Event B? P(B) + P(B ) = 1. Therefore, P(B ) = =.65 = 65% 20. Given Event A, Event B, and Event C. Event A and Event B are mutually exclusive. Event A and Event C are not mutually exclusive. P(A) = 0.55 P(B) = 0.35 P(C) = 0.66 What is the probability of the complement of Event A C? P(A C) = P(A)P(C) =.55 x.66 =.363. Thus, the probability of the complement is = What is the probability that a person will flip a coin three times and that it will land heads up three times in a row? We know that the probability of the coin coming up heads once is.5. So, we multiply.5 x.5 x.5 = A friend offers to play a game where you pay him $2 if the roll of a 6-sided die comes up at 1, 2, 3, or 4, and he pays you $3 if the die comes up a 5 or 6. What is the expected value of a round for you if you play the game? We find the probability of each possible outcome, multiply it by the dollar amount, and then add it together to get the expected value. So, we have (4 x (-2 x 1/6)) + (3 x (3 x 1/6)) = - 1/3.

6 23. Below is a spinner. What is the probability of landing on a number divisible by 2? We divide the number of desired events by the number of possible outcomes to get 4/8 = Using the spinner below, what is the probability of landing on a number divisible by 2 twice in a row? We divide the number of desired events by the number of possible outcomes to get.5 for landing on a number divisible by 2 once. Then, we multiply.5 x.5 =.25 to get the probability of landing on a number divisible by 2 twice. 25. If there is an event in which three decisions must be made and there are four choices available for each decision, how many potential outcomes are there? By the Fundamental Principle of Counting, we multiply the number of choices available to each decision together: 4 x 4 x 4 = 64 possible outcomes

7 26. How many potential combinations can be made from a lottery ticket with four digits, the first two digits being any letter from the alphabet and the last two being any number 0-9. By the Fundamental Principle of Counting, we have 26 x 26 x 10 x 10 = 67,600 possible outcomes. 27. Explain the difference between dependent and independent events. In an independent event, the choice made for a particular decision does not affect the choices available to any other decision. If the event is dependent, then (for at least one decision) making a decision affects the choices available to (at least) one other decision. 28. If we remove all the kings and queens from a 52 card deck, how many unique four card sequences can we draw? First, subtract 8 from 52 to account for the missing kings and queens. Then, using the Fundamental Principle of Counting, we have the potential number of outcomes equal to 44 x 43 x 42 x 41 = 3,258,024 possible outcomes. 29. What classifies a permutation? In a permutation, the order of the elements matters. Therefore, a different ordering of the same elements counts as a different permutation. 30. How many three digit permutations can be made from the numbers 1, 2, 3, 4, and 5 if repetition is not allowed? Since repetition isn t allowed, we use the equation n! / (n-r)! = 5!/(5-3)! = 5!/2! = 60 permutations. 31. How many three digit permutations can be made from the numbers 1, 2, 3, 4, and 5 if repetition is allowed? Since repetition is allowed, we use the equation n r = 3 5 = 243 permutations

8 32. How many possible codes can be formed with a locker that asks for four numbers, numbers cannot be used more than once, and any number can be 0-9? This locker code does not allow repetition. Therefore, we use n! / (n-r)! = 10! / (10-4)! = 10! / 6! = 5,040 permutations. 33. What classifies a combination? In combinations, the order of the elements doesn t matter. Therefore, different orderings of the same elements are considered the same combination. 34. How many combinations without repetition are possible if we have four initial choices for three decisions? This is a combination without repetition, so we use n! / r!(n-r)! = 4! / (3!)(4-3)! = 4! / (3!)(1!) = 4 combinations. 35. How many combinations with repetition are possible if we have four initial choices for three decisions? This is a combination with repetition, so we use (n+r-1)! / r! (n-1)! = 6! / (3!)(3!) = 20 combinations. 36. If Becca is at the store and can buy any three fruit (the store sells apples, oranges, pears, bananas, and kiwis), how many combinations of fruit can she choose? This is an example of a combination with repetition allowed. We use (n+r-1) / r! (n-1!) = 7! / (3!)(4!) = 35 combinations.

9 37. Given the same number of elements and choices to make, which will there be more of: permutations or combinations? Why? There will be more permutations. Because order matters in permutations, there are naturally more permutations given the same number of elements. This is easily identified in the equation for combinations without repetition, which is exactly the same as the one for permutations except we divide by r! 38. If we want to make a five letter word using any letters in the alphabet, how many possible words can we make (a word counts as any sequence of letters in this case)? Feel free to leave your answer in exponential or factorial form. This is a permutation because the order of the letters matters. Repetition is allowed because we can use any letter in the alphabet. Therefore, P(n,r) = n r = 26 5 combinations. 39. How many sandwiches can we make with one slice of turkey, one slice of ham, one slice of cheese, one dollop of mayonnaise, one dollop of ketchup, and one dollop of mustard if we are only using four ingredients? This is a combination because the order of the ingredients doesn t matter. Repetition is not allowed because we have a limited supply of each ingredient. Therefore, C(n,r) = n! / r! (n-r)! = 6! / (4!)(2!) = 15 combinations. 40. How many sandwiches can we make with access to unlimited turkey, ham, cheese, mayonnaise, ketchup, and mustard if we are only using four ingredients? This is a combination because the order of the ingredients doesn t matter. Repetition is allowed because we have an unlimited supply of each ingredient. Therefore, C(n,r) = (n+r-1)! / r! (n-1)! = 9! / (4!)(5!) = 126 combinations.