2. How many ways can the letters in PHOENIX be rearranged? 7! = 5,040 ways.

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1 Math 142 September 27, How many ways can 9 people be arranged in order? 9! = 362,880 ways 2. How many ways can the letters in PHOENIX be rearranged? 7! = 5,040 ways. 3. The letters in MATH are re-arranged. How many of them start with the letter H? The first letter is fixed, leaving three other letters to re-arrange. Hence, 3! = 6 ways. 4. An password contains 4 letters (all lower case, all 26 are allowed) followed by 4 numbers (digits 0-9). How many passwords could be created? (26)(26)(26)(26)(10)(10)(10)(10) = 4,569,760,000. This is the multiplication principle. 5. A race consists of 10 runners. a) How many ways can 1 st -2 nd -3 rd place be filled? (10)(9)(8) = 720. This is also P(10,3). b) Of the remaining seven runners, the next 4 get a ribbon. How many ways can the ribbons be handed out? Since order does not matter, use combination: C(7,4) = 35 ways. 6. A computer byte is a string of 8 digits consisting of 0s and 1s only, for example, is a byte. How many possible bytes are there? 2 =256. Here, use the multiplication principle. 7. From a group of 8 people, a committee of 5 is to be formed. How many committees are possible? Since a committee does not imply order matters, use combination: C(8,5) = From a group of 10 men and 12 women, a committee of 3 men and 4 women is to be formed. How many committees are possible? Since order does not matter, use combinations: Men: C(10,3) = 120, women: C(12,4) = 495. There will be (120)(495) = 59,400 possible committees. 9. Evaluate!. Hint: your calculator may freeze. Try simplifying first.!! = =10,100. Don t depend on your calculator to do big factorials like this. It will! freeze. Instead, reduce as shown, and cancel. 10. How many ways can three six-sided dice be rolled? (6)(6)(6) = 216 ways. (Mult. Princ.) 11. If two dice are rolled, how many of the rolls show a sum of 9? 4 of them do. Hint: write out all 36 outcomes and count off the ones that sum to 9. Writing out the entire sample space is a very big hint, if you catch my drift. 12. If repeats are allowed, you use the multiplication principle. If repeats are not allowed and order matters, you use permutation. If repeats are not allowed and order does not matter, you use combination.

2 MAT 142 Intro to Probability Worksheet (Oct. 4, 2011) 1. A jar of M&M candies contains 12 brown, 4 yellow, 2 blue, 5 red, 3 green and 4 orange. You select one at random. Find the probability that you select one that is: (Leave answers as fractions) a. Brown: 12/30 b. Green or orange: 7/30 c. Not red: 25/30 d. Yellow and blue: 0. It s impossible to select one candy that is both colors simultaneously. Read the questions carefully. And indicates simultaneously, and or indicated union. 2. Shoppers at a local department store were asked to complete a survey of their shopping experience. The results are shown in the table below: Satisfied Not satisfied Total Made a purchase Did not make a purchase Totals a. What is the probability that a shopper selected at random made a purchase? 624/1522. Reduction is not necessary. b. What are the odds that a shopper selected at random was satisfied with the service? 845:677 Must be in odds form! 3. In a study of fatal car accidents, some accidents were attributed to high speed or drunk driving. The probability that the fatality was attributed to high speeds was 0.57 The probability that the fatality was attributed to drunk driving was 0.62 The probability that high speed and drunk driving caused the accident was 0.41 Hint: Make a Venn Diagram a. What is the probability that a fatal accident will be attributed to high speeds or drunk driving? This is the union. b. What is the probability that a fatal accident will be attributed to neither cause? 0.22 c. What is the probability that the fatality will be attributed to drunk driving, but not high speeds? If a single card is selected from a standard 52 card deck, what is the probability that a red face card is selected? 26/52 5. If a single card is drawn from a standard 52 card deck, what is the probability that we obtain a face card or a red card? P(Red) = 26/52, P(Face) = 12/52. P(Both) = 6/52. Therefore, P(Red or Face) = P(Red) + P(Face) P(Both) =26/ /52 6/52 = 32/52.

3 6. If 5 cards are drawn from a standard 52 card deck, what is the probability that the result is 3 red cards and 2 black cards? (, ) (, ) (, ) =0.325, about a 32.5% probability. 7. If the odds for E occurring is 2:5, what is the probability of E? 2/7. Remember, Odds for is written success:failure. Probability is success/total. There are 2 ways to succeed, 5 ways to fail, 7 total. Also, use correct notation. Fractions in place of the colon for odds will be marked wrong! As will the other way around. 8. If a fair coin is tossed three times, write down the sample space S. Then find the probability that exactly two heads occur. S = {ttt, tth, tht, thh, htt, hth, hht, hhh}, P(2 heads) = 3/8 9. If the probability of E occurring is 0.26, what is the probability E does not occur? = 0.74 Look for key words that indicate an ordering or not. Look for phrases like with repetition, which means items can be re-used, or without repetition, which means items cannot be re-used. The formulas for permutation and combination are given but know how to find them on your calculator too. Know how to reduce a large factorial expression. Some calculators will freeze. Always be on the lookout for the words or indicating union, and and indicating intersection (both). A lot of you are still making rookie mistakes by misreading the problems and not picking up on these fine details. If a problem asks for probability, leave your answer in fraction or decimal form. If a problem asks for odds, leave the answer in the colon format. Read the question carefully: odds for is success:failure, odds against is failure:success. Probability is always success/total and they are related by failure + success = total. Draw Venns, Tables or Trees as needed. They help a lot. In conditional probabilities, remember to reduce the sample space accordingly. Read the questions carefully. Look for words like given to indicate a conditional probability. READ ALL QUESTIONS SLOWLY AND CAREFULLY! This is not a race.

4 Math 142 October 11, Two candidates, Smith and Wilson, are running for mayor. The voting breakdown is shown below in the table: Rep (R) Dem (D) Indep (I) Total Smith (S) Wilson (W) Total A voter is selected at random. Determine these probabilities: a) The probability the voter voted for Smith, given the voter was Republican. ( ):72/118 b) The probability the voter was Independent, given the voter voted for Wilson. ( ) 22/129 c) Determine ( ): 68/129. d) Determine ( ): 110/ A jar has 15 red, 20 orange and 22 blue candies. Two candies are drawn without replacement. Find these probabilities: a) The second candy is blue given the first was red. 22/56 b) The second is orange given the first was blue. 20/56 c) Both candies were blue. 22/57 times 21/56 = or 11/76 (both are the same). d) (2 pts extra credit) Both candies are of different color. It s easier to figure the probability of getting the same color first: Two blues is (from part c). Two reds is 0.066, and two oranges is The probability all are the same color is the sum: = Therefore, the probability the two candies are different color is = Tourists to Las Vegas are surveyed. 52% visit Hoover Dam, 31% visit the Strip, and 14% visit both the Strip and Hoover Dam. Determine the following probabilities. You may leave your answer in decimal format. Hint: draw a Venn. a) The probability a tourist visited the Hoover Dam given the tourist visited the Strip. 0.14/0.31 = 0.45 b) The probability a tourist visited Strip given the tourist visited Hoover Dam. 0.14/0.52 = 0.27 c) The probability a tourist did not visit Hoover Dam given the tourist did not visit the Strip. The probability a tourist did not visit the strip is The probability the tourist visited neither place is Thus, the probability is 0.31/0.69 = 0.45

5 4. You roll a single die once. If it lands a 6, you get $10. Otherwise, you get nothing. The cost to play is free. What is the expected value of one roll of this die? There is a 1/6 chance of winning $10, 5/6 chance of winning nothing. The EV is (1/6)(10) + (5/6)(0) = 10/6 = $ A bag has 20 tokens in it. They all feel the same. One is gold colored and worth $20. Two are silver colored and worth $5 each. The other 17 are worth nothing. For $3, you can reach in and randomly grab one token. What is the expected value of this game? Subtract out the $3 cost when figuring the winnings: You have a 1/20 chance of netting $17, a 2/20 chance of netting $2, and a 17/20 chance of losing your $3. The EV is (1/20)(17) + (2/20)(2) + (17/20)(-3) = -30/20 = This means on average, you ll lose $1.50 per game. In the long term, it s a bad game. 6. A lottery sells 100 tickets for $1 each. One ticket is the winner, with a jackpot of $75. The rest are worthless, and you lose your $1. Your friend s bright idea is to buy all the tickets. Use Expected Value to explain why this is a lousy idea. Show your calculation and give a one sentence explanation. The EV is (1/100)(74) + (99/100)(-1) = -25/100 = -$0.25. The EV is negative so you ll lose in the long term. If you spent $100 to get $75, you have lost $ A roulette wheel has 38 slots. The cost to play is $1. If the ball lands in a slot you picked, you win $36. Otherwise, you lose the $1. a) Find the Expected Value of one play. (1/38)(35) + (37/38)(-1) = -2/38 = -$ Is this game in your favor? (Y/N) No. You ll lose on average a little over five cents per game in the long term. b) If you played 100 games, how much up or down can you expect to be? 100 times the EV; (100)(- $0.053) = -$5.30. You ll be down about $5.30. c) What is the fair price to play this game? Fair price = cost + EV = $1 + (-$0.053) = $0.947, or about 95 cents.

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