1 / 39

Sec. 1.6 Probability

Sec. 1.6 Probability. Objective: Students set up probability equations appropriately. Experimental Probability. Probability of event = Number of times event occurs Number of trials. Example 1.

lareina-garrett
Download Presentation

Sec. 1.6 Probability

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Sec. 1.6 Probability Objective: Students set up probability equations appropriately

  2. Experimental Probability • Probability of event = • Number of times event occurs Number of trials

  3. Example 1 • A player hit the bull’s eye on a circular dartboard 8 times out of 50. Find the experimental probability that the player hits the bull’s eye.

  4. Number of times event occurs = Number of trials We need to use the formula.

  5. Example 2 • Find the theoretical probability of rolling a multiple of 3 with a number cube? How about rolling an odd? • The Cube is a normal six sided di.

  6. A) How many numbers on the cube are a multiple of 3? • Yes 2 numbers, 3 and 6. • So we get. 2 = 1 6 3 B) How many odds? Yes 3 numbers, 1,3,5 So we get 3 = 1 6 2

  7. Example 3 • Suppose that all the points on the circular dartboard shown below are equally likely to be hit by a dart you have thrown. Find the probability of only scoring 2 points with one throw. • Note: The radius of each circle is one unit larger than the one below it. 20 20 10 5 2

  8. First we need to find the area of the whole dart board. This is the denominator because any throw can hit any where on the dart board. • To find the area of the green we need to subtract the areas of the others. So we get (using area πr2 of a circle) • π(4r)2 – π(3r)2 π(4r)2 =16πr2 - 9πr2 16πr2 = 7πr2 16πr2 20 20 10 5 2

  9. 6.7a Warm-Up Evaluate the expression • 3! 2) 7! • 3) ( 4!/6!) = 3 • 2 • 1 = 7 • 6 • 5 • 4 • 3 • 2 • 1 = 6 = 5040 = (4 • 3 • 2 • 1) / (6 • 5 • 4 • 3 • 2 • 1) = 1/(6 • 5) = 1/30

  10. 6.7a Permutations and Combinations Objective–To be able to use permutations to count the number of ways an event can happen State Standard – 18.0 Students use the fundamental counting principles to compute combinations and permutations. 19.0Students use Combinations and Permutations to compute probabilities.

  11. White White White Ham Wheat Wheat Wheat Rye Rye Rye Turkey Roast Beef White Pastrami on White Pastrami Wheat Pastrami on Wheat Rye Pastrami on Rye The Fundamental Counting Principle Suppose you own a small deli. You offer 4 types of meat (ham, turkey, roast beef, and pastrami) and 3 types of bread (white, wheat, and rye). How many choices do your customers have for a meat sandwich? Because you have 4 meats and 3 breads: The Total # of choices is 4 • 3 = 12

  12. 4th digit 5th digit 6th digit 7th digit 3rd digit • 10 • 10 • 10 • 10 • 10 Example 1a At a restaurant, you have a choice of 8 different entrees, 2 different salads, 12 different drinks, and 6 different desserts. How many different dinners consisting of 1 salad, 1 entrée, 1 drink, and 1 dessert can you choose? 8 • 2 • 12 • 6 = 1152 different dinners Example 1b How many different 7 digit phone numbers are possible if the first digit cannot be 0 or 1? 2nd digit 1st digit 8 • 10 = 8,000,000

  13. Permutations of n objects taken r at a time The number of permutations of r objects taken from a group of n distinct objects is denoted by nPr and is given by: nPr = n!/(n – r)! A Permutation – is an ordering of n objects. For instance, there are 6 permutations of the letters A, B, and C: ABC, ACB, BAC, BCA, CAB, CBA In general there are 3 choices for the 1st letter, 2 choices for the 2nd letter, and 1 choice for the 3rd letter so there are: 3 • 2 • 1 =6 ways to arrange the letters. n!

  14. Example 2 Find the number of permutations. 5 P2 nPr = n!/(n – r)! 5 P2 = 5!/(5 – 2)! = 5 • 4• 3• 2• 13 • 2• 1 = 20 = 5 • 4

  15. 6.7b Warm-Up Find the Product: • (2x – 3)2 2) (x – 3y)2 = (2x – 3)(2x – 3) = (x – 3y)(x – 3y) = 4x2 – 12x + 9 = x2 – 6xy + 9y2

  16. 6.7b Permutations and Combinations Objective–To be able to use combinations to count the number of ways an event can happen State Standard – 19.0Students use Combinations and Permutations to compute probabilities.

  17. Combinations of n objects taken r at a time The number of combinations of r objects taken from a group of n distinct objects is denoted by nCr and is given by: nCr = n! (n – r)! • r!

  18. Example 3a Find the number of combinations. 5 C2 nCr = n!/{(n – r)! • r!} 5 C2 = 5!/{(5 – 2)! • 2!} = 5 • 4• 3! 3!• 2! = 5 • 4 2 = 10

  19. Example 3b Find the number of combinations. 10 C4 nCr = n!/{(n – r)! • r!} 10 C4 = 10!/{(10 – 4)! • 4!} = 10 • 9 •8 • 7• 6! 6!• 4! 3 = 10 • 9•8 • 7 4 • 3•2 • 1 = 210

  20. 52! 52C7 = (52 – 7)! • 7! Example 4 Use a standard deck of 52 cards. a) If the order is not important, how many different 7-card hands are possible? =133,784,560 b) How many of these hands have exactly 6 cards of the same suit? 4C1 • 13C6 = 6,864 When finding the # of ways both an event A and an event B can occur you need to multiply.

  21. 9.7 Warm-Up A bag contains 24 green marbles, 22 blue marbles, 14 yellow marbles, and 12 red marbles. Suppose you pick one marble at random. Find each probability. • P(yellow) 2) P(not blue) • 3) P(green or red)

  22. 9.7 Probability of Multiple Events Objective–To be able to find the probabilities of events A and B, and of events A or B. State Standard – 1.0 Students will know the definition of the notion of independent events and can use the rules of addition, multiplication, and complementation to solve for probabilities of particular events in finite sample spaces.

  23. Dependent Events – Is when the outcome of one event affects the outcome of a second event. Independent Events – Is when the outcome of one event does not affect the outcome of a second event. Example 1 Classify each pair of events as dependent or independent. a) Toss a coin. Then, select a marble from a bag that contains marbles of different colors. Independent b) Select a marble from a bag that contains marbles of two different colors. Put the marble aside, and select a second marble from the bag. Dependent

  24. Probability of A and B If A and B are independent events, the probability that both A and B occur is: P(A and B) = P(A)·P(B) Checking for Understanding: Suppose you select a marble from a bag of marbles. You replace the marble and then select again. Are your selections dependent or independent event? Explain. Independent; The number of marbles is the same after the marble is replaced.

  25. Example 2: A random number generator on a computer selects three integers from 1 to 20. What is the probability that all three numbers are less than or equal to 5? The probability of selecting a number from 1 to 5 is: So the probability that all three numbers will be less than or equal to 5 is:

  26. Example 3: A person draws 2 cards from a deck of 52 cards. What is the probability that the 2 cards drawn will both be face cards if the deck contains 12 face cards? = 0.05

  27. Probability 12.2 Conditional Probability 8.4.1 MATHPOWERTM 12, WESTERN EDITION

  28. Conditional Probability If A and B are events from an experiment, the conditional probability of B given A (P(A|B)), is the probability that Event B will occur given that Event A has already occurred. The conditional probability is equal to the probability that B and A will occur divided by the probability that B will occur. This is given in Bayes’ Formula: 8.4.2

  29. Conditional Probability Determine the conditional probability for each of the following: a) Given P(B and A) = 0.725 and P(B) = 0.78, find P(A|B). P(A|B) = 0.9295 • Given P(blonde and tall) = 0.5 and P(blonde) = 0.73, • find P(A|B). P(A|B) = 0.6849 8.4.3

  30. yes no female 8 6 male 5 7 Do you own a pet? 8 14 Therefore, P(own a pet | female) equals . Conditional Probability LESSON 12-2 Additional Examples The table shows the results of a class survey. Find P(own a pet | female) The condition female limits the sample space to 14 possible outcomes. Of the 14 females, 8 own a pet.

  31. Material Recycled Non-Recycled Paper 36.7 45.1 Metal 6.3 11.9 Glass 2.4 10.1 Plastic 1.4 24.0 Other 21.2 70.1 24.0 45.1 + 11.9 + 10.1 + 24.0 + 70.1 P(plastic | non-recycled) = 24.0 161.2 = 0.15 Conditional Probability LESSON 12-2 Additional Examples Using the data in the table, find the probability that a sample of non-recycled waste was plastic. The given condition limits the sample space to non-recycled waste. A favorable outcome is non-recycled plastic. The probability that the non-recycled waste was plastic is about 15%.

  32. Relate: P( male ) = 58% P( male and jogs ) = 20% Define: Let A = male. Let B = jogs. P( A and B ) P( A ) Write: P( A | B ) = 0.2 0.58 = Substitute 0.2 for P(A and B) and 0.58 for P(A). 0.344 Simplify. Conditional Probability LESSON 12-2 Additional Examples Researchers asked people who exercise regularly whether they jog or walk. Fifty-eight percent of the respondents were male. Twenty percent of all respondents were males who said they jog. Find the probability that a male respondent jogs. The probability that a male respondent jogs is about 34%.

  33. On 28% of the days, the sky was mostly clear. During the mostly clear days, it rained 4% of the time. During the cloudy days it rained 31% of the time. Conditional Probability LESSON 12-2 Additional Examples A student made the following observations of the weather in his hometown. P(rain|clear) P(norain|clear) P(rain|cloudy) P(norain|cloudy) Use a tree diagram to find the probability that a day will start out clear, and then it will rain.

  34. Conditional Probability LESSON 12-2 Additional Examples (continued) The path containing clear and rain represent days that start out clear and then will rain. P(clear and rain) = P(rain | clear) • P(clear) = 0.04 • 0.28 = 0.011 The probability that a day will start out clear and then rain is about 1%.

  35. Mean: x = 6+7+9+9+9+10 = 50 = 8.33 6 6 12.4 Warm Up Find the mean, median and mode of the data set. 1) 6, 7, 9, 9, 9,10 Median: 9 Mode: 9

  36. 12.4 Standard Deviation

  37. Standard Deviation standard deviation- is often very important in scientific experiments. A small standard deviation signifies the results are consistent. A large standard deviation signifies the results are inconclusive. The standard deviation (read as “sigma”) of x1,x2…. xn  = (x1 – x )2+(x2 – x )2+…+(xn - x)2 n

  38. Mean: x = 86 + 89 + 91 = 88.67 3 Example 1 What is the S.D. of the 3 test scores? 86, 89, and 91  = (x1 – x )2+(x2 – x )2+…+(xn - x)2 n  = 2.06

  39. Mean: x = 70 + 85 + 99 + 40 = 294 = 73.5 4 4 Example 2 What is the S.D. of the 4 test scores? 70, 85, 99 and 40  = (x1 – x )2+(x2 – x )2+…+(xn - x)2 n  = 21.89

More Related