Probability of Mutually Exclusive Events.

1. Probability of Mutually Exclusive Events. • A ~ You are at school • B ~ You are bowling • A∩B ~ You are at school and bowling • A∩B = Ø (you can’t bowl at school) S

2. 2 Sets are disjoint if their intersection is the empty set Ø • Disjoint Sets B e.g. There is no bowling alley at school so you can not both be bowling and at school.

3. Probability: • P(A U B)=P(A)+P(B) • In the school and bowling example we can simply add the probabilities. • We are not concerned with overlap because these events are: “Mutually Exclusive”. • The equation above only works when events are mutually exclusive.

4. EXAMPE • You are at school 7x5=35 hours out of 168 hours in a week. P( at School)=35/168=21% • You bowl for 2 hours once a month p( bowling) = 2/(24x30)= 0.003 or 0.3% • P( School U bowling)= 21% + 0.3% = 21.3%

5. Text Book • Your text book refers to the set A U B as A or B. This is the union of two sets and is meant to be interpreted as one or the other or both. When sets A and B are mutually exclusive we can remove the ‘both’ condition from the sentence above.

6. When Events are NOT Mutually Exclusive • Just as with counting when we subtract the number of elements in the intersection: n( A UB) = n( A) + n( B) – n( A∩B ) • We subtract the probability of the intersection: P( A UB) = P( A) + P( B) – P( A∩B )

7. EXAMPLE 2 • You spend 7 hours a week having lunch, 5 of those lunch hours are at school. • What’s P( At School U Eating Lunch) ?

8. EXAMPLE 2 • You spend 7 hours a week having lunch, 5 of those lunch hours are at school. • What’s P( At School U Eating Lunch) ? • A ~ At School B ~ Eating Lunch • P(A)=35/168=21% P(B)=7/168=4% • P( A∩B ) = 5/168=3% • P(AUB) = P(A)+P(B)-P(A∩B) = 21% + 4% - 3% = 22% We subtract P(A∩B) so that we don’t count the intersection twice.

9. n( A UB) = n( A) + n( B) – n( A∩B ) • e.g. A = set of DM Students B = set of P.E. Students There are 28 DM students and 25 PE students. 12 are in both. How many students in total are in DM or PE (or both)? n(A)=28, n(B)=25, n( A∩B )=12 n( A UB) = n( A) + n( B) – n( A∩B ) =28 + 25 – 12 = 41

10. Example 3 Of the 120 students in a class, 30 speak Chinese, 50 speak Spanish, 75 speak French, 12 speak Spanish and Chinese, 30 speak Spanish and French, and 15 speak Chinese and French. Seven students speak all three languages. What is the probability that a randomly chosen student speaks none of these languages?

11. 7 speak all 3 15 speak Chinese and French 15 30 speak Spanish and French 12 speak Spanish and Chinese 23 5 50 speak Spanish 7 10 30 speak Chinese 37 8 75 Speak French 120 – 105 = 15 don’t speak any.

12. P( a random student doesn’t speak any) • = 15/120 • =12.5 % • We used a Venn Diagram because it is better suited for tracking the intersections of sets in this case.

13. Adding Intersections With an Event and it’s compliment. • Often when dealing with more than one event it is useful to use: • P(A) = P(A∩B)+P(A∩B’) • Example: Only parts that are not obviously defective get shipped. Consider events: • G~part is good, OD~part is obviously defective, OD’ ~ part is not obviously defective • P(G) = P(G∩OD)+P(G∩OD’) • P(G) = 0+P(G∩OD’) • P(G) = P(G∩OD’) (this intersection can now be used to calculate conditional probability.

14. Example 3.14 (an old test question) • 11)You are trying to find a popular toy for your little sister for Christmas. If you go to Toys R Them, there is a 66% probability of finding it. If you go to FloorMart there is an 80% chance of you finding it. • a) If you only have time to go to one store and you decide which one by tossing a coin, what is the probability that you find the toy for your sister? (4 marks) setTheoryProbExample.notebook

15. b) If you have a 60% chance of going to Toys R Them and also have a 60% chance of going to FloorMart (regardless of whether or not you went to Toys R Them) what is the probability that you will find the toy for your sister? (4 marks) setTheoryProbExample.notebook

16. Homework • Page 340 Questions 1 to 9