1. Chapter three Conditional Probability and Independence
STAT 111
Chapter three
Conditional Probability and Independence

2. Conditional Probability
The probability of an event A occurring when it is known that some event A has occurred is called a conditional probability and is denoted by P(A\B). The symbol P(A\B) is usually read The probability that A occurs given that B occurs or simply The probability of A given B.
Definition The conditional probability of A, given B, denoted by P(A\B), is defined

3. Example 1
A coin is flipped twice. What is the conditional probability that both flips result in heads, given that the first flip does?
S = {HH, HT, TH, TT} , n(S) = 4
Let A: the event both flips result in heads A= {HH} , n(A) = 1
B: the event the first flip results in head B ={ HH ,HT}, n(B) = 2 A ∩ B = { HH}, n(A ∩ B ) = 1
P (A ∩ B ) = 1 / 4
P ( B ) = 2 / 4
P (A \ B ) =P (A ∩ B ) /P ( B ) = 1/2

4. Example 2 Let A and B be events with P(A)=1/2,and P(B)=1/3,
total AC A
1/3
1/12
1/4 B
2/3
5/12 BC 1
1/2
Example 2
Let A and B be events with P(A)=1/2,and P(B)=1/3,
and P(A ∩ BC)=1/4. Find
P(A\B)
P(A ∩ B)=P(A)- P(A ∩ BC)=1/2-1/4=1/4 (or from table)
P(A\B)= P(A ∩ B)/P(B)=(1/4)/(1/3)=3/4
2. P(B\A)
P(B\A)= P(A ∩ B)/P(A)=(1/4)/(1/2)=2/4
3. P(BC|AC)=P(AC ∩ BC)/ P(AC)=(5/12)/(1/2)=5/6

5. Example 3
In a certain college, 25% of the students failed Mathematics, 15% of the students failed chemistry and 10% of the students failed both mathematics and chemistry. A student is selected at random, find
If he failed chemistry, what is the probability that he failed Mathematics? P(M)=0.25, P(C)=0.15, P(M∩C)=0.1
If the failed mathematics, what is the probability he failed chemistry?

6. Example 4
If A and B are disjoint events and P(B)>0, What is the value of P(A\B)? A and B are disjoint P(A ∩ B)=0 P(A \ B) = P(A ∩ B)/P(B)=0/P(B)=0

7. P(.\B) is a Probability
Conditional probabilities satisfy all of the properties of ordinary probabilities. This is proved by the following proposition, which shows that P(A\B) satisfies the definition of probability. Proposition (a) 0 ≤ P(A \ B) ≤ 1 (b)P(S\B)=1 (c) If A1, A2 ,… are mutually disjoint sets in S , then

8. Notes Note that for any event A and B where P(B)>0 P(Φ \B)= 0
P(AC\B)=1-P(A\B)
P(A\B)=P(A∩C\B)+P(A∩Cc\B)
P(AUC\B)=P(A\B)+P(C\B)-P(A∩C\B)
If A  C then P(A|B)≤ P(C\B).

9. Independence
As stated in chapter 2, two events A and B are independent if the occurrence or nonoccurrence of either of them has no relation to the occurrence or nonoccurrence of the other, that is the probability that both A and B will occur is equal to the product of their individual probabilities which implies P(A ∩ B) = P(A)P(B)
Theorem
If A and B are independent, then ,
P(A \ B) =P(A) if P(B)>0 and
P(B \ A)=P(B) if P(A)\>0

10. Example Given P(A)=0.5, and P(A U B)=0.6, find P(B \ A) if
A and B are independent
P(AUB ) = P(A) + P(B)- P(A∩ B) = P(A) + P(B)- P(A)P(B)
P(AUB ) = P(A) + P(B)(1- P(A))
0.6= P(B)
P(B)=0.2
P(B \ A)= P(B)=0.2
A and B are disjoint
A and B are disjoint P (A ∩ B)=0
P(B \ A)= 0

11. Multiplicative Rules
As previously defined, the conditional probability of A given B is P(A \ B) = P (A ∩ B)/P ( B ) if P (B) > 0 From which P(A∩B)=P(B)P(A\B) This is called the multiplicative rule which helps to calculate the probability that two events will both occur. This rule is often used when the two events A and B are not independent. A generalization of the multiplicative rule which provides an expression for the probability of the intersection of an arbitrary number of events, is the following P(A1∩ A2…An)=P(A1)P(A2\ A1)P(A3 \ A1 ∩ A2 )…P(An \ A1 ∩… An-1)

12. Example 1
Suppose that a fuse صمامه كهربائية))box containing 20 fuses, of which 5 are defective. If 2 fuses are selected at random and removed from the box in succession متتابعة ) ) without replacing the first, what is the probability that both fuses are defective? a1 first fuse is defective a2 second fuse is defective P(A1∩ A2)=P(A1)P(A2\ A1)=5/20 x 4/19 =1/19 Or using counting

13. Example 2 b1 drawing a black ball from bag 1
One bag contains 4 white balls and 3 black balls, and a second bag contains 3 white balls and 5 black balls. One ball is drawn form the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from the second bag is black?
b1 drawing a black ball from bag 1
B2 drawing a black ball from bag 2
W1 drawing a white ball from bag 1
 P ( B 2) = P ( B2 ∩ B1) + P ( B2 ∩ W1)
= P (B1)P(B2\ B1 ) + P (W1) P (B2 \ W1)
=3/7 x 6/9 + 4/7 x 5/9 =0.6

14. Bayes' Theorem
Let S denote the sample space of some experiment, and consider k events A1, A2,..., Ak are disjoint and . It is said that these events form a partition of S. If the k events A1, A2,..., Ak form a partition of S and if B is any other event in S, then the events A1 ∩ B,A2 ∩ B,..,,Ak ∩ B will form a partition of B, Hence, we can write B = (A1∩B)U(A2∩B)U… U (Ak∩B) Furthermore, since the k events on the right side of this equation are disjoint, Finally, if P(Ai)>0 for i=1 ,2,...,k then P(Ai∩B)=P(Ai)P(B\Ai ) and it follows that

15. Bayes' Theorem
Bayes' Theorem Let the events A1, A2,..., Ak form a partition of the space S such
that P(Aj)>0, for j=1,..,k,
and let B be any event such that P(B)>0. Then, for j=1,..., k,

16. Example 1
Three machines A, B and C produce respectively 60%, 30%, and 10% of the total number of items of a factory. The percentages of defective output of these machines are respectively 2%, 3%, and 4%.
Solution:
Let D denote the output is defective, A denote item from machine A, B denote item from machine B and C denote item from machine C.
P (A) = P (B) = P (C ) = 0.1
P ( D \ A) = P ( D \ B) = P (D \ C ) = 0.04
1. An item is selected at random and is found defective, what is the probability that the item was produced by machine C.
2. An item is selected at random and is found defective, what is the probability that the item was produced by machine C or B.
3. An item is selected at random and is found defective, what is the probability that the item was not produced by machine C.

17. Example 2
Three cooks, A, B and C bake a special kind of cake, and with respective probabilities 0.02, 0.03, and 0.05 it fails to rise. In the restaurant where they work, A bake 50 percent of these cakes, B 30 percent and C 20 percent. What proportion of failures is caused by A {P(A|F)}?
Solution
Let F: that cake fail to rise
P(A)= P(B)= P(C)=0.2
P(F|A)= P(F|B)= P(F|C)=0.05

18. Note
Note that if P(A1),....,P(Ak) are not given then we assume that these events are
equally likely each P(Ai) = 1/k