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1 Events and Their Probabilities Folks often want to know the probability of an event so they can plan their day (or whatever)

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Presentation on theme: "1 Events and Their Probabilities Folks often want to know the probability of an event so they can plan their day (or whatever)"— Presentation transcript:

1 1 Events and Their Probabilities Folks often want to know the probability of an event so they can plan their day (or whatever)

2 2 Remember from a previous section that an experiment has sample points. I think it is good to say that each sample point is mutually exclusive. In other words, on any given trial of the experiment one and only one sample point can result (it could be different for each subject, but each subject has only one). An event is a specific collection of sample points or experimental outcomes. An event is said to occur if any one of the specific sample points results. Example: Say when you go to a vending machine and buy a bag of M&M’s, the bag has 40 pieces inside and the colors in the bag are red, brown, blue, orange, yellow and green (I can not remember the real colors and amount anymore - I have not purchased them lately !) An event might be when you open the bag, the first color out is yellow. Another event might be the first color out is red or blue.

3 3 (Bells and whistles are blaring because the following is an important idea.) The probability of any event is equal to the sum of the probabilities of the sample points in the event. Say by the relative frequency approach we know in the M&M example that P(red) = 4/40, P(brown) = 10/40, P(blue) = 8/40, P(orange) = 8/40, P(yellow) = 4/40 and P(green) = 6/40. A special event is the sample space. In other words, what is the probability that the first one out is red, brown, blue, orange, yellow or green? The probability here is 4/ /40 + 8/40 + 8/40 + 4/40 + 6/40 = 1.

4 4 What is the probability the first M&M out of the bag is not brown? Not brown means red, blue, orange, yellow, or green. So, P(not brown) = P(reb) + P(blue) + P(orange) + P(yellow) + P(green) = 4/40 + 8/40 + 8/40 + 4/40 + 6/40 = 30/40 =.75. Say event A is the situation that either of my two favorite M&M’s comes out first - red or yellow. Also say event B is that either of your two favorite M&M’s comes out - red or blue. P(A) = P(red) + P(yellow) = 4/40 + 4/40 = 8/40 =.2 P(B) = P(red) + P(blue) = 4/40 + 8/40 = 12/40 =.3

5 5 Now, in the last example on the previous slide, red was part of both events A and B. Both events A and B would occur if red occurred. The intersection of events is itself an event that has sample points that are in both the original events. So, the intersection of A and B is red and P(red) =.1 The union of events is also an event, but it is an event that lists once each sample point from either of the original events. The union of A and B would have red, yellow and blue as the sample points involved and the probability is P(red) + P(yellow) + P(blue) = =.4

6 6 Back on slide three I had the following sentence. The probability of any event is equal to the sum of the probabilities of the sample points in the event. In theory, this method can always be used to find the probability of an event. In fact, I have really relied on it up to this point. But, sometimes this method is difficult to apply and so we use some other rules of probability to assist us in finding the probability of an event. The Complement of an Event Say event A is well defined. The complement of A is the event consisting of all points not in A and is denoted A c. By definition, P(A) + P(A c ) = 1. This may be useful to find the P(A) because P(A) = 1 - P(A c ), or to find P(A c ) = 1 - P(A). What is the probability the first M&M out of the bag is not brown? P(not Brown) = 1 - P(Brown) = =.75

7 7 Unions and Intersections An event can be a “mixture” of other events. Say we have events A and B. The union of events A and B, written A ∪ B, is an event that includes all sample points belonging to A or B. The intersection of events A and B, written A ∩ B, is an event that includes all sample points belonging to both A and B. A BI have used two circles to represent the events, as Mr. Venn did with his diagram many moons ago. Note here that A and B overlap. All the points inside the box represent all the outcomes of the experiment.

8 8 Note on the previous screen I write The union of events A and B The intersection of events A and B I write A and B, but the context of the problem will require the use of the union or the intersection. The union really means is the experimental outcome in either A or B. When you look at the Venn Diagram, the union is combing all of circle A and all of circle B. But, with the overlap in my diagram, you would not include this overlap area twice. The intersection really means is the experimental outcome in both A and B. The intersection is just the overlap I mentioned above with the union.

9 9 Say event A is the situation that either of my two favorite M&M’s comes out first - red or yellow. Also say event B is that either of your two favorite M&M’s comes out - red or blue. Now, using the idea that the probability of any event is equal to the sum of the probabilities of the sample points in the event, we note P(A) = P(red) + P(yellow) = =.2, and P(B) = P(red) + P(blue) = =.3 Now P(A ∩ B) = P(red) =.1, because red is the only one in both.

10 10 Before I showed P(A ⋃ B) = P(red) + P(yellow) + P(blue) = =.4 But, in general we write P(A ⋃ B) = P(A) + P(B) - P(A ∩ B). This is the additional law for the union of events. In the example both P(A) and P(B) include P(red), but since we only want to include it once we subtract out the intersection. P(A ⋃ B) = =.4.

11 11 If two events are mutually exclusive, then the events have no overlap. This means that P(A ∩ B) = 0, and thus the union of mutually exclusive events is just P(A) + P(B). Say E1, E2, E3, E4, and E5 are all equally likely outcomes and we have events A, B, and C such that A={E1, E2}, B={E3, E4} and C={E2, E3, E5}. Then a. P(A) = P(E1) + P(E2) = =.4 (remember The probability of any event is equal to the sum of the probabilities of the sample points in the event.), P(B) = P(E3) + P(E4) =.4, and P(C) = P(E2) + P(E3) + P(E5) =.6 b. P(A ∪ B) = P(A) + P(B) =.8, since A and B are mutually exclusive. c. A c = {E3, E4, E5} and P(A c ) = 1 - P(A) =.6, C c = {E1, E4}, P(C c ) = 1 - P(C) =.4. d. P(A υ B c ) = =.6 e. P(B ∪ C) = =.8


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