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# Qa06 probability

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### Qa06 probability

1. 1. QA06 PROBABILITY 9/11/2013 1
2. 2. 9/11/2013 2
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6. 6. 1) tree diagram 2) sample space 3) event 4) Fundamental Counting Principle 5) Factorial Counting Outcomes  Count outcomes using a tree diagram.  Count outcomes using the Fundamental Counting Principle.
7. 7. Counting Outcomes One method used for counting the number of possible outcomes is to draw a tree diagram. The last column of a tree diagram shows all of the possible outcomes. The list of all possible outcomes is called the sample space, while any collection of one or more outcomes in the sample space is called an event.
8. 8. Counting Outcomes A football team uses red jerseys for road games, white jerseys for home games, and gray jerseys for practice games. The team uses gray or black pants, and black and white shoes. Use a tree diagram to determine the number of possible uniforms.
9. 9. Counting Outcomes A football team uses red jerseys for road games, white jerseys for home games, and gray jerseys for practice games. The team uses gray or black pants, and black and white shoes. Use a tree diagram to determine the number of possible uniforms. Jersey Pants Shoes Outcomes Red White Gray
10. 10. Counting Outcomes A football team uses red jerseys for road games, white jerseys for home games, and gray jerseys for practice games. The team uses gray or black pants, and black and white shoes. Use a tree diagram to determine the number of possible uniforms. Jersey Pants Shoes Outcomes Red White Gray Gray Black Gray Black Gray Black
11. 11. Counting Outcomes A football team uses red jerseys for road games, white jerseys for home games, and gray jerseys for practice games. The team uses gray or black pants, and black and white shoes. Use a tree diagram to determine the number of possible uniforms. Jersey Pants Shoes Outcomes Red White Gray Gray Black Gray Black Gray Black Black White Black White Black White Black White Black White Black White
12. 12. Counting Outcomes A football team uses red jerseys for road games, white jerseys for home games, and gray jerseys for practice games. The team uses gray or black pants, and black and white shoes. Use a tree diagram to determine the number of possible uniforms. Jersey Pants Shoes Outcomes Red White Gray Gray Black Gray Black Gray Black Black White Black White RGB RGW RBB RBW WGB WGW WBB WBW GGB GGW GBB GBW Black White Black White Black White Black White
13. 13. Counting Outcomes A football team uses red jerseys for road games, white jerseys for home games, and gray jerseys for practice games. The team uses gray or black pants, and black and white shoes. Use a tree diagram to determine the number of possible uniforms. Jersey Pants Shoes Outcomes Red White Gray Gray Black Gray Black Gray Black Black White Black White RGB RGW RBB RBW WGB WGW WBB WBW GGB GGW GBB GBW Black White Black White Black White Black White The tree diagram shows that there are 12 possible outcomes.
14. 14. Counting Outcomes In the previous example, the number of possible outcomes could also be found by multiplying the number of choices for each item.
15. 15. Counting Outcomes In the previous example, the number of possible outcomes could also be found by multiplying the number of choices for each item. The team could choose from: 3 different colored jerseys 2 different colored pants 2 different colored shoes
16. 16. Counting Outcomes In the previous example, the number of possible outcomes could also be found by multiplying the number of choices for each item. The team could choose from: 3 different colored jerseys 2 different colored pants 2 different colored shoes There are 3 X 2 X 2 or 12 possible uniforms.
17. 17. Counting Outcomes In the previous example, the number of possible outcomes could also be found by multiplying the number of choices for each item. The team could choose from: 3 different colored jerseys 2 different colored pants 2 different colored shoes There are 3 X 2 X 2 or 12 possible uniforms. This example illustrates the Fundamental Counting Principle.
18. 18. Counting Outcomes In the previous example, the number of possible outcomes could also be found by multiplying the number of choices for each item. The team could choose from: 3 different colored jerseys 2 different colored pants 2 different colored shoes There are 3 X 2 X 2 or 12 possible uniforms. This example illustrates the Fundamental Counting Principle. If an event M can occur in m ways, and is followed by an event N that can occur in n ways, then the event M followed by event N can occur m X n ways.
19. 19. Counting Outcomes A deli offers a lunch special in which you can choose a sandwich, a side dish, an a beverage. If there are 10 different sandwiches, 12 different side dishes, and 7 different beverages, from which to choose, how many different lunch specials can be ordered?
20. 20. Counting Outcomes A deli offers a lunch special in which you can choose a sandwich, a side dish, an a beverage. If there are 10 different sandwiches, 12 different side dishes, and 7 different beverages, from which to choose, how many different lunch specials can be ordered? Multiply to find the number of lunch specials. sandwich choices side dish choices beverage choices number of specialsX X =
21. 21. Counting Outcomes A deli offers a lunch special in which you can choose a sandwich, a side dish, an a beverage. If there are 10 different sandwiches, 12 different side dishes, and 7 different beverages, from which to choose, how many different lunch specials can be ordered? Multiply to find the number of lunch specials. sandwich choices side dish choices beverage choices number of specialsX X = 10 X 12 X 7 = 840 The number of different lunch specials is 840.
22. 22. Counting Outcomes A.J. is setting up a display of the ten most popular video games from the previous week. If he places the games side-by-side on a shelf, in how many different ways can he arrange them?
23. 23. Counting Outcomes A.J. is setting up a display of the ten most popular video games from the previous week. If he places the games side-by-side on a shelf, in how many different ways can he arrange them? The number of ways to arrange the games can be found by multiplying the number of choices for each position.
24. 24. Counting Outcomes A.J. is setting up a display of the ten most popular video games from the previous week. If he places the games side-by-side on a shelf, in how many different ways can he arrange them? The number of ways to arrange the games can be found by multiplying the number of choices for each position.  A.J. has 10 games from which to choose for the first position.
25. 25. Counting Outcomes A.J. is setting up a display of the ten most popular video games from the previous week. If he places the games side-by-side on a shelf, in how many different ways can he arrange them? The number of ways to arrange the games can be found by multiplying the number of choices for each position.  A.J. has 10 games from which to choose for the first position.  After choosing a game for the first position, there are nine games left from which to choose for the second position.
26. 26. Counting Outcomes A.J. is setting up a display of the ten most popular video games from the previous week. If he places the games side-by-side on a shelf, in how many different ways can he arrange them? The number of ways to arrange the games can be found by multiplying the number of choices for each position.  A.J. has 10 games from which to choose for the first position.  After choosing a game for the first position, there are nine games left from which to choose for the second position.  There are now eight choices for the third position.
27. 27. Counting Outcomes A.J. is setting up a display of the ten most popular video games from the previous week. If he places the games side-by-side on a shelf, in how many different ways can he arrange them? The number of ways to arrange the games can be found by multiplying the number of choices for each position.  A.J. has 10 games from which to choose for the first position.  After choosing a game for the first position, there are nine games left from which to choose for the second position.  There are now eight choices for the third position.  This process continues until there is only one choice left for the last position.
28. 28. Counting Outcomes A.J. is setting up a display of the ten most popular video games from the previous week. If he places the games side-by-side on a shelf, in how many different ways can he arrange them? The number of ways to arrange the games can be found by multiplying the number of choices for each position.  A.J. has 10 games from which to choose for the first position.  After choosing a game for the first position, there are nine games left from which to choose for the second position.  There are now eight choices for the third position.  This process continues until there is only one choice left for the last position. Let n represent the number of arrangements. n = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 or 3,628,800
29. 29. Counting Outcomes A.J. is setting up a display of the ten most popular video games from the previous week. If he places the games side-by-side on a shelf, in how many different ways can he arrange them? The number of ways to arrange the games can be found by multiplying the number of choices for each position.  A.J. has 10 games from which to choose for the first position.  After choosing a game for the first position, there are nine games left from which to choose for the second position.  There are now eight choices for the third position.  This process continues until there is only one choice left for the last position. Let n represent the number of arrangements. n = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 or 3,628,800 There are 3,628,800 different ways to arrange the video games.
30. 30. Counting Outcomes The expression n = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 used in the previous example can be written as 10! using a factorial. The expression n!, read n factorial, where n is greater than zero, is the product of all positive integers beginning with n and counting backward to 1. n! = n(n – 1)* (n – 2) * . . . 3 * 2 * 1 Example: 5! = 5 * 4 * 3 * 2 * 1 or 120
31. 31. Counting Outcomes 0! is defined as being equal to 1. Let’s see why.
32. 32. Counting Outcomes 0! is defined as being equal to 1. Let’s see why. 4 !4 !3 Writing this out using the definition of factorials. 4 1*2*3*4 4 1*2*3*4 !3
33. 33. Counting Outcomes 0! is defined as being equal to 1. Let’s see why. 4 !4 !3 Writing this out using the definition of factorials. 4 1*2*3*4 4 1*2*3*4 !3 3 !3 !2
34. 34. Counting Outcomes 0! is defined as being equal to 1. Let’s see why. 4 !4 !3 Writing this out using the definition of factorials. 4 1*2*3*4 4 1*2*3*4 !3 3 !3 !2 2 !2 !1
35. 35. Counting Outcomes 0! is defined as being equal to 1. Let’s see why. 4 !4 !3 Writing this out using the definition of factorials. 4 1*2*3*4 4 1*2*3*4 !3 3 !3 !2 2 !2 !1 1 !1 !0so, the next logical conclusion is that
36. 36. 9/11/2013 40 An unbiased die is tossed. Find the probability of getting a multiple of 3? The possible options are : 1 to 6. there are only 2 multiples of 3 : 3,6 so probability is (number of favourable outcomes ) / (total number of possibilities) = 2/6 = 1/3 answer
37. 37. 9/11/2013 41 In a simultaneous throw of a pair of dice,find the probability of getting a total more than 7? We can have 36 possibilities (6*6) however, we need only those cases where the total is 8 or more. These are : (6,2),(6,3),(6,4),(6,5),(6,6),(5,3),(5,4),(5,5),(5,6),(4,4),(4,5),(4,6 ),(3,5),(3,6),(2,6) =15 answer = 15/36 = 5/12 answer
38. 38. 9/11/2013 42 A bag contains 6 white and 4 black balls .Two balls are drawn at random .Find the probability that they are of the same colour? Both are white : 6/10*5/9 both are black = 4/10*3/9 add them : =42/90 or 7/15 or : 6c2/10C2*1/2 + 4c2/10c2 =21/45 = 7/15 answer
39. 39. 9/11/2013 43 Two dice are thrown together.What is the probability that the sum of the number on the two faces is divisible by 4 or 6? The possibilities are : (1,3)(1,5) (2,2) (2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6) thus we are able to get 14 out of 36. so answer = 7/18 answer
40. 40. 9/11/2013 44 Two cards are drawn at random from a pack of 52 cards What is the probability that either both are black or both are queens? Both are black = 26/52 * 25/51=25/102 both are queens : 4/52 * 3/51=3/663 both are black queens : 2/52*1/51 = 1/1326 now add them : (25/102 + 3/663 – 1/1326) =(325+6-1)/1326 =330/1326 or .25 answer
41. 41. 9/11/2013 45 Two dices are tossed the probability that the total score is a prime number? Prime numbers are : 1,2,3,5,7,11 totals are : (1,2),(1,1),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4 ),(4,1),(4,3),(5,2),(5,6),6,1),(6,5) =15/36 answer
42. 42. 9/11/2013 46 Two dice are thrown simultaneously .what is the probability of getting two numbers whose product is even? If any one of the two numbers is an even number, the product will be even number. Thus we should pick up all those cases when both the numbers are odd numbers : (1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3) (5,5) thus there are only 9 such cases. Remove them from 36, we get : 27 cases answer : 27/36 answer
43. 43. 9/11/2013 47 In a lottery ,there are 10 prizes and 25 blanks. A lottery is drawn at random. what is the probability of getting a prize ? 10/(10+25) =10/35 or 2/7 answer
44. 44. 9/11/2013 48 In a class ,30 % of the students offered English,20 % offered Hindi and 10 %offered Both. If a student is offered at random, what is the probability that he has offered English or Hindi? 30+20-10 = 40% or .4 answer
45. 45. 9/11/2013 49 Two cards are drawn from a pack of 52 cards .What is the probability that either both are Red or both are Kings? Both are red ½ * 25/51 both are king = 4/52 + 3/51 now add both these answers =55/221
46. 46. 9/11/2013 50 one card is drawn at random from a pack of 52 cards.What is the probability that the card drawn is a face card? Face cards are : Jack, queen, king total = 12 12/52 answer
47. 47. 9/11/2013 51 A man and his wife appear in an interview for two vacancies in the same post.The probability of husband's selection is 1/7 and the probabililty of wife's selection is 1/5.What is the probabililty that only one of them is selected? Husband + not wife =1/7 * 4/5 = 4/35 wife + not husband =1/5 * 6/7 = 6/35 add = 10/35 answer
48. 48. 9/11/2013 52 From a pack of 52 cards,one card is drawn at random.What is the probability that the card is a 10 or a spade? 4/52 + 13/52 – 1/52 =16/52 answer
49. 49. 9/11/2013 53 A bag contains 4 white balls ,5 red and 6 blue balls .Three balls are drawn at random from the bag.What is the probability that all of them are red ? 5/15*4/14*3/13 or 5c2/15c2 = =2/91
50. 50. 9/11/2013 54 A box contains 10 block and 10 white balls.What is the probability of drawing two balls of the same colour? Both are black : 10/20 * 9/19 =9/38 +both are white : 10/20 * 9/19 =9/38 or black : 10c2 / 20c2 +white : 10c2 / 20c 2 =90/190
51. 51. 9/11/2013 55 A box contains 20 electricbulbs ,out of which 4 are defective, two bulbs are chosen at random from this box.What is the probability that at least one of these is defective ? In such questions (at least one type), it is better to reverse the question, solve it and deduct the answer from 1. So here we shall first calculate the probability of getting no defective bulb. Let us assume that no bulb is defective : 16/20 * 15/19 = 12/19 at least one is defective = 1 -12/19 = 7/19 answer
52. 52. 9/11/2013 56 Two cards are drawn together from apack of 52 cards.What is the probability that one is a spade and one is a heart ? First is spade and 2nd heart : 13/52 * 13/51 = 13/204 First is heart and 2nd spade : 13/52 * 13/51 = 13/204 add them : 13/102 answer
53. 53. 9/11/2013 57 The probability that a card drawn from a pack of 52 cards will be a diamond or a king? 13/52 + 4/52 – 1/52 =16/52

May. 7, 2015

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