Download presentation

Presentation is loading. Please wait.

Published byLuis Rubey Modified about 1 year ago

1
Probability Three basic types of probability: Probability as counting Probability as geometry Probability as Algebra

2
Probability as counting

3
Example #1 Find the probability that when two standard 6-sided dices are rolled, the sum of the numbers on the top faces is 5. # of total outcomes: 6 * 6 /2 = 18 (regardless of sequence) # of successful outcomes: 1,4; 2,3 (regardless of sequence) Answer = 2 / 18 = 1/9

4
Example #2 A bag contains 16 marbles, 4 of which are blue and 12 of which are green. Two marbles are randomly pulled from the bag at the same time. What is the probability that both marbles are blue? # of total outcomes: 16*15/2 = 120 (regardless of sequence) # of successful outcomes: 4*3/2 = 6 (regardless of sequence) Answer = 6 / 120 = 1/20

5
Probability as geometry

6
Example #3 Lawrence parked his car in a parking lot at a randomly chosen time between 2:30PM and 4:00 PM. Exactly half an hour later he drove his car out of the parking lot. What is the probability that he left the parking lot after 4:00 PM? Answer = 1/3

7
Example #4 If 1 ≤ x ≤ 4 and 2 ≤ y ≤ 6, find the probability that x + y ≥ 5. For 1 ≤ x ≤ 4, 2 ≤ y ≤ 6, we can draw: For x + y = 5, we can draw:

8
Example #4 Area of triangle: 2*2 * ½ = 2 Area of shaded: 3*4 – 2 = 10 Total area: 3 * 4 = 12 Answer = (area of shaded)/(total area) = 10/12 = 5/6

9
Probability as algebra If you can identify enough relationships among those related probabilities, you might consider solve the problem algebraically.

10
Example #5 If the probability that it rains next Tuesday in Seattle is twice the probability that it doesn’t, what is the probability that it rains next Tuesday in Seattle? Let X be the probability to rain on Tuesday We conclude: X + X/2 = (1) (Note: Total probability is always 1) Solve X from (1), we get: X = 2/3

11
Example #6 Johnny and Michael play a game in which they take turns rolling a pair of fair dice until one of them rolls “snake eyes” (both dice show 1’s). That person is the winner. If Johnny goes first, what is the probability that Johnny wins the game? Probability for J won in first round is: 1/36 Probability for J won in second round is: (35/36) * (35/36) * (1/36) Probability for J won in third round is: (35/36) * (35/36) * (35/36) * (35/36) * (1/36)

12
Example #6 Johnny and Michael play a game in which they take turns rolling a pair of fair dice until one of them rolls “snake eyes” (both dice show 1’s). That person is the winner. If Johnny goes first, what is the probability that Johnny wins the game? Probability for J to win is: 1/36 + 1/36 * (35/36) 2 + 1/36 * (35/36) 4 + 1/36 * (35/36) 6 + … Let S = 1 + (35/36) 2 + (35/36) 4 + (35/36) 6 + … We get S = 1 + (35/36) 2 * (1 + (35/36) 2 + (35/36) 4 + …) Hence S = 1 + (35/36) 2 * S Solve for S, we can get: S = 36 * 36 / 71 Thus the answer = (1/36) * S = 36/71

13
Basic Principal of Probability Probability of compound events: (i.e. a series of events) If a compound event E that can be completed in three steps, A, B, and C, and the probability to complete each step is a, b, and c, then the probability of E is a * b * c.

14
Basic Principal of Probability Probability of exclusive events: (or alternative approaches) If a task E can be achieved via three different approaches A, B, and C, each with probability of success of a, b, and c. Then the probability of E to be successful is a + b + c.

15
Example #7 Four green cards and three red cards are placed in a bag and selected at random. 1) What is the probability of selecting two red cards without replacement? 2) What is the probability of selecting a green card then a red card with replacement? 3) What is the probability of selecting three cards of the same color without replacement?

16
Example #7 Four green cards and three red cards are placed in a bag and selected at random. 1) What is the probability of selecting two red cards without replacement? 2) What is the probability of selecting a green card then a red card with replacement? 3) What is the probability of selecting three cards of the same color without replacement? 1) Two steps to get two red cards. Prob. of 1 st step: 3/7; prob. of 2 nd step: 2/6 Thus the answer: 3/7 * 2/6 = 1/7 2) Two steps to get two red cards. Prob. of 1 st step: 4/7; prob. of 2 nd step: 3/7 Thus the answer: 4/7 * 3/7 = 12/49 3) Prob. of 3 red: 3/7 * 2/6 * 1/5 Prob. of 3 green: 4/7 * 3/6 * 2/5 Thus the answer: 4/35 + 1/35 = 5/35 = 1/7

17
Example #7 Four green cards and three red cards are placed in a bag and selected at random. 1) What is the probability of selecting two red cards without replacement? 2) What is the probability of selecting a green card then a red card with replacement? 3) What is the probability of selecting three cards of the same color without replacement? 1) Total # of outcome: 7*6/2 = 21 Desired # of outcome: 3*2/2 = 3 Answer: 3/21 = 1/7 Alternatively, the problem can also be solved via counting. 2) Total # of outcome: 7*7/2 Desired # of outcome: 4*3/2 Answer: 12/49 3) Total # of outcome: 7*6*5 / (3*2*1) = 35 Desired # of outcome: (g) 4!/3! + (r) 3!/3! = 5 Answer: 5/35 = 1/7

18
Compound events and counting combined Example #8 Three fair dice are rolled. What is the probability that exactly two of the rolls show a 1? For one dice, prob. of showing 1 is: 1/6 For 11X (X ≠ 1) the prob. is 1/6 * 1/6 * 5/6 Note that 11X, 1X1, and X11 have equal prob. Thus the answer = 3 * (1/6) * (1/6) * (5/6) = 5/72

19
Probability and combination When objects are put into different groups, consider to use the formula: (# of desired group) * (# of groups to pick) (# of total groups)

20
Probability and combination Example #9 digits 1 through 9 are divided into three sets of three digits. What is the probability that the product of one of the sets is odd? Total # of groups: C(9,3) = (9*8*7) / (3*2) = 84 # of desired groups (groups with odd #s only): (5*4*3) / (3 * 2) = 10 Thus the answer = (10 / 84) * 3 = 5/14 There are 3 groups to be formed

21
Complimentary counting & probability When too many cases needs to be counted, consider use complimentary counting, and subtract the result from 1.

22
Complimentary counting & probability Example #10 Three coins are flipped. What’s the probability of flipping at least one heads and at least one tail? Total # of outcome: 2 3 = 8 # of cases for all heads: 1 Thus the answer = 1 – ¼ = 3/4 Prob. of all heads or all tails: 2/8 = ¼ # of cases for all tails: 1

23
Expected Values Expected values = Average value of all possible outcome

24
Expected Values Example #11: There are $1 and $5 bills in a bag. The expected value for a randomly selected bill is $1.20. What is the fewest # of bills that could be in the bag? Let x be # of $1 bills, and y be # of $5 bills. The total value is x + 5y, and # of bills is x + y Thus the smallest # for x is 1, and smallest # for y is 19. The fewest # of bills is x + y = 20. Thus we have (x + 5y) / (x + y) = 1.2; and we get 3.8 y = 0.2 x; or y/x = 19/1

25
Conditional probability Let P a&b = probability of event A & B occurs Let P b = probability of event B occurs Let P a/b = probability of event A if event B occurs P a&b P a/b = P b

26
Conditional probability Example #12: Suppose I have two cards, one with a blue side and a red side, the other with two red sides. I choose one card at random, and placed it on the table. The side on top is red. What’s the probability that the other side is also red? Total # of cases (with red top-face): 3 ( 1 with blue-red card, 2 with red-red card.) Thus the answer = 2/3 # of desired cases: 2 (under the condition that top face is red)

27
Conditional probability Example #12: Suppose I have two cards, one with a blue side and a red side, the other with two red sides. I choose one card at random, and placed it on the table. The side on top is red. What’s the probability that the other side is also red? Probability of red top & bottom: 1/2 Probability of red on top: 3/4 Conditional probability: ½ / ¾ = 2/3

28
Alternative counting Example #13: A fair coin is to be tossed ten times. Find the probability that heads never occur on consecutive tosses. Let S n = # of desired outcome with n toss. Observe that a tail must occur in one of the first two toss. If the first toss is tail, then there are S n-1 outcomes from the next n-1 tosses. If the first toss is head, then there are S n-2 outcomes from the next n-2 tosses. And we have S n = S n-1 + S n-2

29
Alternative counting Example #13 A fair coin is to be tossed ten times. Find the probability that heads never occur on consecutive tosses. Observe that: S 1 = 2; S 2 = 3; we get: S 3 = = 5 S 4 = = 8 S 5 = 5+8 = 13 … S 10 = S 8 + S 9 = = 144 Total # of outcome: 2 10 Answer = 144 / 2 10 = (2 4 * 3 * 3)/2 10 = 9/64

30
Alternative counting Example #13 A fair coin is to be tossed ten times. Find the probability that heads never occur on consecutive tosses. Total # of outcome: 2 10 Now count the desired outcome: Observe that there must be at least 5 tails. Otherwise there will be at least 6 heads, and at least two of them will be next to each other. Consider the case there are 5 tails, then the 5 heads must be in any of the “ _” slots under _ T _ T _ T _ T _ T _. There are 6 slots for 5 heads. The total # is: C(6, 5) = (6*5*4*3*2) / 5! = 6

31
Alternative counting Example #13 A fair coin is to be tossed ten times. Find the probability that heads never occur on consecutive tosses. Consider the case there are 6 tails, then the 4 heads must be in any of the “ _” slots under _ T _ T _ T _ T _ T _ T _. There are 7 slots for 4 heads. The total # is: C(7, 4) = (7*6*5*4) / 4! =35 Continue this pattern, we got: # of desired outcome = C(6,5) + C(7,4) + C(8,3) + C(9,2) + C(10, 1) + C(11,0) = = 144 Answer = (# of desired outcome) / (total # outcome) = 144 / 2 10 = 9/64.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google