1. Some concepts of probabilities: set, experiment, sample space, event, …

2. Complement: A and Ac

and: P(A) + P(Ac)=1. 3. Venn Diagram

In the easiest case, we first count the number of outcomes in the sample space, assuming each outcome is equally likely, then count the number of outcomes for the event of interest, do the division.

Example: Event A={Roll a fair die and get at least 4} Sample space {1,2,3,4,5,6} , 6 outcomes Outcomes for this event: {4,5,6}, 3 outcomes P(A)=3/6=0.5.

From Venn diagram, we deal with the probability associated with two or more events.

Using two events, A and B as an example. We may be interested in A happens, B

happens, A or B happens, A and B happens, A happens but B does not, B happens but A does not.

A or B happens, P(A U B), it is called union of A and B

Both A and B happens, P(A∩B), it is called the intersection of A and B.

Addition law: P(A U B) = P(A) + P(B) - P(A∩B).

A class of 30 students took 2 midterms during a semester. 22 of them passed the first one and 26 of them passed the second one. If 3 students failed both, find the number of students who passed the each of the two midterms and failed the other.

Answer: 21 passed both, 1 passed the first but failed the second, 5 passed the second but failed the first, 3 failed both.

Think about the midterm example: A={Pass midterm 1}; B={Pass midterm 2} P(A∩B) = ? P(A∩B) ={ Pass both exam 1 and 2} P(A U B)=? P(A U B) ={ Pass at least ONE exam }

Think about P(A+B) or P(AUB)= P(A)+P(B)- P(A∩B) Or, probability that one passes at least one

exam = probability that one passes midterm 1 + probability that one passes midterm 2 – probability that one passes both.

Since pass midterm 1 includes 2 possibilities pass midterm 1 and pass midterm 2;

pass midterm 1 and fail midterm 2.So does pass midterm 2.

Therefore P(AUB)=P(A)+P(B)- P(A∩B) =22/30+26/30-21/30 =27/30=0.9

A survey of 100 college students revealed the following facts about their sporting hobbies. Let S be the event that a person swims, R be the event that a person runs, and B be the event that a person plays baseball. Create a Venn Diagram for the following data.

  42 run 34 play baseball 13 only swim 6 swim and run 14 swim and play baseball 11 run and play baseball 22 do no sport 2 swim, run, and play baseball

Let A={Run}, B={Baseball} and C={Swim} 42 run {A} 34 play baseball {B} 13 only swim {Ac ∩ Bc ∩ C }

6 swim and run {A ∩ C} 14 swim and play baseball {C ∩ B} 11 run and play baseball {A ∩ B} 22 do no sport {Ac ∩ Bc ∩ Cc} 2 swim, run, and play baseball {A∩ B ∩ C}

Calculate P(A+B+C)By addition law:P(A+B+C)=P(A)+P(B)+P(C)- P(A∩B)- P(A∩C)-

P(B∩C)+ {A∩ B ∩ C}From the given information we know

everything except P(C). But P(C) can be figured out from the Venn Diagram.

Therefore, P(A+B+C)=(42+34+31-6-14-11+2)/100=78/100

And it confirms what we found in Venn Diagram

P(AUB) c =P(Ac ∩ Bc ) Or (AUB) c = (Ac ∩ Bc ) It simply says, the complement of the union

of two events is the intersection of the complement of the two events.

For example: in the midterm example (AUB) means pass either of two midterms (AUB) c means does not pass either of two

midterms It is the same as saying fail both, or (Ac ∩ Bc )

Also, think about the sports example: {A+B+C} means playing at least one of

running, baseball or swimming. According to Demorgans Law, (AUBUC ) c = (A c ∩ B c ∩ C c ) And (A c ∩ B c ∩ C c ) is given in the problem,

which is 22, Therefore, {A+B+C} =100- (AUBUC )

c=100- (A c ∩ B c ∩ C c ) =100-22=78