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A card is drawn from a standard deck of 52 cards. Find each probability.

1. ANSWER. 2. 1. ANSWER. 13. Lesson 10.4 , For use with pages 707-713. A card is drawn from a standard deck of 52 cards. Find each probability. 1. P (a red card). 2. P (a ten). 1. ANSWER. 4. Lesson 10.4 , For use with pages 707-713.

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A card is drawn from a standard deck of 52 cards. Find each probability.

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  1. 1 ANSWER 2 1 ANSWER 13 Lesson 10.4, For use with pages 707-713 A card is drawn from a standard deck of 52 cards. Find each probability. 1. P(a red card) 2. P(a ten)

  2. 1 ANSWER 4 Lesson 10.4, For use with pages 707-713 A card is drawn from a standard deck of 52 cards. Find each probability. 3. What is the probability that two flipped coins show heads?

  3. Probability Dependent Events10.4

  4. Let event Abe selecting a 10 and event Bbe selecting a face card. Ahas 4 outcomes and Bhas 12 outcomes. Because Aand Bare disjoint, the probability is: 4 4 16 12 0.308 = = + = 13 52 52 52 EXAMPLE 1 Find probability of disjoint events A card is randomly selected from a standard deck of 52 cards. What is the probability that it is a 10or a face card? SOLUTION P(A orB) = P(A)+ P(B)

  5. SOLUTION Let event Abe selecting a face card and event Bbe selecting a spade. Ahas 12 outcomes and Bhas 13 outcomes. Of these, 3 outcomes are common to Aand B. So, the probability of selecting a face card or a spade is: EXAMPLE 2 Standardized Test Practice

  6. = P(A)+ P(B)– P(AandB) P(A orB) 12 13 3 – = + 52 52 52 22 = 52 11 = 26 ANSWER The correct answer is B. EXAMPLE 2 Standardized Test Practice

  7. Senior Class Out of 200 students in a senior class, 113 students are either varsity athletes or on the honor roll. There are 74 seniors who are varsity athletes and 51 seniors who are on the honor roll. What is the probability that a randomly selected senior is both a varsity athlete and on the honor roll? EXAMPLE 3 Use a formula to find P(A and B) SOLUTION Let event Abe selecting a senior who is a varsity athlete and event Bbe selecting a senior on the honor roll. From the given information you know that

  8. 113 51 74 P(A) ,P(B) , andP(A orB) = = = 200 200 200 P(A orB) = P(A)+P(B)–P(A andB) 74 51 113 –P(A andB) + = 200 200 200 74 51 113 P(A andB) – + = 200 200 200 12 3 P(A andB) = = = 0.06 200 50 EXAMPLE 3 Use a formula to find P(A and B) FindP( A and B ) Write general formula. Substitute known probabilities. Solve for P(Aand B). Simplify.

  9. Selecting an ace or an eight 2 4 2 4 = = + 13 52 13 52 ANSWER for Examples 1, 2, and 3 GUIDED PRACTICE A card is randomly selected from a standard deck of 52 cards. Find the probability of the given event. SOLUTION Let event Abe selecting an ace and event Bbe selecting an eight. Ahas 4 outcomes and B also has 4 outcomes. Because Aand Bare disjoint, the probability is: P(A orB) = P(A)+ P(B)

  10. A card is randomly selected from a standard deck of 52 cards. Find the probability of the given event. 1 52 Selecting a 10or a diamond 4 16 13 4 = + – = = 52 52 52 13 0.308 ANSWER for Examples 1, 2, and 3 GUIDED PRACTICE SOLUTION Let event Abe selecting a 10 and event Bbe selecting a diamond. Ahas 4 outcomes and Bhas 13 outcomes. Of these, 1 outcome is common to A and B. So the probability is: P(A orB) = P(A)+ PB – P(Aand B)

  11. 51 32 P(B) P(A) = = 200 200 What If? In Example 3, suppose 32 seniors are in the band and 64 seniors are in the band or on the honor roll. What is the probability that a randomly selected senior is both in the band and on the honor roll? 64 200 for Examples 1, 2, and 3 GUIDED PRACTICE SOLUTION Let event Abe selecting a senior who is a varsity athlete and event Bbe selecting a senior on the honor roll. From the given information you know that , andP(A orB) =

  12. ANSWER 0.095 P(A andB) = P(A orB) = P(A)+P(B)–P(A andB) 32 51 64 –P(A andB) + = 200 200 200 32 51 64 P(A andB) – + = 200 200 200 P(A andB) 19 = = 83 – 64 200 200 for Examples 1, 2, and 3 GUIDED PRACTICE FindP( A and B ) Write general formula. Substitute known probabilities. Solve for P(Aand B). Simplify.

  13. Dice When two six-sided dice are rolled, there are 36 possible outcomes, as shown. Find the probability of the given event. The sum is not 6. The sum is less than or equal to 9. EXAMPLE 4 Find probabilities of complements

  14. P(sum is not 6) = 1 – P(sum is 6) 5 = 1 – 36 31 0.861 = 36 P(sum< 9) = 1 – P(sum > 9) 6 36 = 1– 30 = 36 5 0.833 = 6 EXAMPLE 4 Find probabilities of complements

  15. Fortune Cookies A restaurant gives a free fortune cookie to every guest. The restaurant claims there are 500 different messages hidden inside the fortune cookies. What is the probability that a group of 5 people receive at least 2 fortune cookies with the same message inside? The number of ways to give messages to the 5 people is 5005. The number of ways to give different messages to the 5 people is 500499498 497496. So, the probability that at least 2 of the 5 people have the same message is: EXAMPLE 5 Use a complement in real life SOLUTION

  16. P(at least2are the same) = 1 – P(none are the same) 500499498 497496 = 1 – 5005 0.0199 EXAMPLE 5 Use a complement in real life

  17. Find P( A ). P(A) = 0.45 P ( A ) = 1–p(A) P( A ) 1 – 0.45 = 0.55 ANSWER = for Examples 4 and 5 GUIDED PRACTICE SOLUTION The probability of the complement of A is

  18. Find P( A ). 3 1 1 P(A) = 4 4 4 P ( A ) = 1–p(A) P( A ) 1 – = ANSWER = for Examples 4 and 5 GUIDED PRACTICE SOLUTION The probability of the complement of A is

  19. Find P( A ). P(A) = 1 P ( A ) = 1–p(A) P( A ) 1 – 1 = 0 ANSWER for Examples 4 and 5 GUIDED PRACTICE SOLUTION The probability of the complement of A is

  20. Find P( A ). P ( A ) = 1–p(A) P(A) = 0.03 P ( A ) 1 – 0.03 = 0.97 ANSWER for Examples 4 and 5 GUIDED PRACTICE SOLUTION The probability of the complement of A is

  21. The number of ways to give messages to the 5 people is 1005. The number of ways to give different messages to the 5 people is 1009998 9796. So, the probability that at least 2 of the 5 people have the same message is: What If? In Example 5, how does the answer change if there are only 100 different messages hidden inside the fortune cookies? for Examples 4 and 5 GUIDED PRACTICE SOLUTION

  22. P(at least2are the same) = 1 – P(none are the same) 1009998 9796 = 1 – 1005 0.097 = ANSWER The probability increases to about 0.097. for Examples 4 and 5 GUIDED PRACTICE

  23. 10.5 Independent and Dependent Probability

  24. EXAMPLE 1 Standardized Test Practice SOLUTION Let events Aand Bbe getting the winning ticket for the gift certificate and movie passes, respectively. The events are independent. So, the probability is:

  25. 5 5 1 1 P(A andB) = P(A)P(B) = = 150 200 30 40 The correct answer is B. ANSWER 1 = 1200 EXAMPLE 1 Standardized Test Practice

  26. What If? In Example 1, what is the probability that you win the mall gift certificate but not the booklet of movie passes? 5 195 P(A andB) = P(A)P(B) = 150 200 ANSWER 1 39 = 30 40 39 = 1200 13 = 400 13 400 for Example 1 GUIDED PRACTICE

  27. Racing In a BMX meet, each heat consists of 8 competitors who are randomly assigned lanes from 1 to 8. What is the probability that a racer will draw lane 8 in the 3 heats in which the racer participates? 1 1 1 P(A andB andC) = P(A)P(B)P(C) = 8 8 8 1 0.00195 = 512 EXAMPLE 2 Find probability of three independent events SOLUTION Let events A, B, and Cbe drawing lane 8 in the first, second, and thirdheats, respectively. The three events are independent. So, the probability is:

  28. Music While you are riding to school, your portable CD player randomly plays 4 different songs from a CD with 16 songs on it. What is the probability that you will hear your favorite song on the CD at least once during the week (5 days)? 15C4 P(not hearing song) = 16C4 EXAMPLE 3 Use a complement to find a probability SOLUTION For one day, the probability of not hearing your favorite song is:

  29. P(hearing song) = 1– [P(not hearing song)]5 ( ) 15C4 5 0.763 = 1– 16C4 EXAMPLE 3 Use a complement to find a probability Hearing or not hearing your favorite song on Monday, on Tuesday, and so on are independent events. So, the probability of hearing the song at least once is:

  30. SPINNER : A spinner is divided into ten equal regions numbered 1 to 10. What is the probability that 3 consecutive spins result in perfect squares? 3 3 3 P(A andB andC) = P(A)P(B)P(C) = ANSWER 0.027 10 10 10 27 0.027 = 1000 for Examples 2 and 3 GUIDED PRACTICE

  31. What If? In Example 3, how does your answer change if the CD has only 12 songs on it? 11C4 P(not hearing song) = 12C4 for Examples 2 and 3 GUIDED PRACTICE SOLUTION For one day, the probability of not hearing your favorite song is: Hearing or not hearing your favorite song on Monday, on Tuesday, and so on are independent events. So, the probability of hearing the song at least once is:

  32. ( ) 11C4 5 = 1– 12C4 P(hearing song) = 1– [P(not hearing song)]5 ( ) 330 5 = 1– 495 ( ) 5 = 1– 0.66 0.1252 = 1– 0.87 = ANSWER It increases to about 0.87. for Examples 2 and 3 GUIDED PRACTICE

  33. Weather The table shows the numbers of tropical cyclones that formed during the hurricane seasons from 1988 to 2004. Use the table to estimate (a) the probability that a future tropical cyclone is a hurricane and (b) the probability that a future tropical cyclone in the Northern Hemisphere is a hurricane. EXAMPLE 4 Find a conditional probability

  34. Number of hurricanes P(hurricane) = Total number of Cyclones 760 0.483 = 1575 P(hurricane Northern Hemisphere) Number of hurricanes Northern Hemisphere = Total number of Cyclones Northern Hemisphere 545 0.477 = 1142 EXAMPLE 4 Find a conditional probability SOLUTION

  35. Selecting Cards You randomly select two cards from a standard deck of 52 cards. What is the probability that the first card is not a heart and the second is a heart if (a) you replace the first card before selecting the second, and (b) you do not replace the first card? EXAMPLE 5 Comparing independent and dependent events SOLUTION Let Abe “the first card is not a heart” and Bbe “the second card is a heart.”

  36. If you replace the first card before selecting the second card, then Aand Bare independent events. So, the probability is: 13 3 39 P(A and B) = P(A) P(B) = 0.188 = 52 16 52 If you do not replace the first card before selecting the second card, then Aand Bare dependent events. So, the probability is: 13 13 39 P(A and B) = P(A) P(B A ) 0.191 = = 51 68 52 EXAMPLE 5 Comparing independent and dependent events

  37. What If? Use the information in Example 4 to find (a) the probability that a future tropical cyclone is a tropical storm and (b) the probability that a future tropical cyclone in the Southern Hemisphere is a tropical storm. Number of tropical storm P(tropical storm) = Total number of Cyclones 598 0.379 = = 1575 0.38 ANSWER for Examples 4 and 5 GUIDED PRACTICE SOLUTION

  38. 200 = 433 = 0.46 P(tropical storm / Southern Hemisphere) 0.46 ANSWER Number of tropical storm in Southern Hemisphere = Total number of Cyclones Southern Hemisphere for Examples 4 and 5 GUIDED PRACTICE

  39. A spade, then a club for Examples 4 and 5 GUIDED PRACTICE Find the probability of drawing the given cards from a standard deck of 52 cards (a) with replacement and (b) without replacement. SOLUTION Let Abe “the first card is a spade” and Bbe “the second card is a club.”

  40. If you replace the first card before selecting the second card, then Aand Bare independent events. So, the probability is: 13 1 13 P(A and B) = P(A) P(B) = = 52 16 52 If you do not replace the first card before selecting the second card, then Aand Bare dependent events. So, the probability is: 13 13 13 P(A and B) = P(A) P(B A ) = = 51 204 52 for Examples 4 and 5 GUIDED PRACTICE

  41. A jack, then another jack for Examples 4 and 5 GUIDED PRACTICE Find the probability of drawing the given cards from a standard deck of 52 cards (a) with replacement and (b) without replacement. SOLUTION Let Abe “the first card is a jack” and Bbe “the second card is a jack.”

  42. If you replace the first card before selecting the second card, then Aand Bare independent events. So, the probability is: 1 4 4 P(A and B) = P(A) P(B) = = 169 52 52 If you do not replace the first card before selecting the second card, then Aand Bare dependent events. So, the probability is: 3 1 4 P(A and B) = P(A) P(B A ) = = 51 221 52 for Examples 4 and 5 GUIDED PRACTICE

  43. Costume Party You and two friends go to the same store at different times to buy costumes for a costume party. There are 15 different costumes at the store, and the store has at least 3 duplicates of each costume. What is the probability that you each choose different costumes? EXAMPLE 6 Find probability of three dependent events SOLUTION Let event Abe that you choose a costume, let event Bbe that one friend chooses a different costume, and let event Cbe that your other friend chooses a third costume. These events are dependent. So, the probability is:

  44. P(A andB andC) = P(A) P(B A) P(C A andB) 14 15 13 = 15 15 15 182 0.809 = 225 EXAMPLE 6 Find probability of three dependent events

  45. Safety Using observations made of drivers arriving at a certain high school, a study reports that 69% of adults wear seat belts while driving. A high school student also in the car wears a seat belt 66% of the time when the adult wears a seat belt, and 26% of the time when the adult does not wear a seat belt. What is the probability that a high school student in the study wears a seat belt? EXAMPLE 7 Solve a multi-step problem

  46. A probability tree diagram, where the probabilities are given along the branches, can help you solve the problem. Notice that the probabilities for all branches from the same point must sum to 1. EXAMPLE 7 Solve a multi-step problem SOLUTION

  47. P(C) = P(A andC) + P(B andC) = P(A) P(C A) + P(B) P(C B) = 0.536 = (0.69)(0.66) +(0.31)(0.26) EXAMPLE 7 Solve a multi-step problem So, the probability that a high school student wears a seat belt is:

  48. for Examples 6 and 7 GUIDED PRACTICE What If? In Example 6, what is the probability that you and your friends choose different costumes if the store sells 20 different costumes? SOLUTION Let event Abe that you choose a costume, let event Bbe that one friend chooses a different costume, and let event Cbe that your other friend chooses a third costume. These events are dependent. So, the probability is:

  49. P(A andB andC) = P(A) P(B A) P(C A andB) 19 20 18 = 20 20 20 = 0.855 = 0.855 ANSWER 1 0.95 0.72 for Examples 6 and 7 GUIDED PRACTICE

  50. P(C) = P(A andC) + P(B andC) BASKETBALL : A high school basketball team leads at halftime in 60% of the games in a season. The team wins 80% of the time when they have the halftime lead, but only 10% of the time when they do not. What is the probability that the team wins a particular game during the season? 8. = P(A) P(C A) + P(B) P(C B) = 0.52 = (0.60)(0.80) +(0.40)(0.10) 52% ANSWER for Examples 6 and 7 GUIDED PRACTICE

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