1. General Probability Rules
Lesson 6 – 3b
General Probability Rules

2. Knowledge Objectives
Define what is meant by a joint event and joint probability
Explain what is meant by the conditional probability P(A | B)
State the general multiplication rule for any two events
Explain what is meant by Bayes’s rule.

3. Construction Objectives
State the addition rule for disjoint events
State the general addition rule for union of two events
Given any two events A and B, compute P(A  B)
Given two events, compute their joint probability
Use the general multiplication rule to define P(B | A)
Define independent events in terms of a conditional probability

4. Vocabulary
Personal Probabilities – reflect someone’s assessment (guess) of chance
Joint Event – simultaneous occurrence of two events
Joint Probability – probability of a joint event
Conditional Probabilities – probability of an event given that another event has occurred

5. General Multiplication Rule
The probability that two events A and B both occur is P(A and B) = P(A  B) = P(A) ∙ P(B | A)
where P(B | A) is a conditional probability read as the probability of B given that A has occurred

6. Conditional Probability Rule
If A and B are any two events, then
P(A and B) N(A and B)
P(B | A) = =
P(A) N(A)
N is the number of outcomes

7. Independence in Terms of Conditional Probability
Two events A and B are independent if P(B | A) = P(B)
Example: P(A = Rolling a six on a single die) = 1/6
P(B = Rolling a six on a second roll) = 1/ no matter what was rolled on the first roll!!
So probability of rolling a 6 on the second roll, given you rolled a six on the first is still 1/6 P(B | A) = P(B) so A and B are independent

8. Contingency Tables Male Female Total Right handed 48 42 90 Left handed12 8 20 60 50
110
What is the probability of left-handed given that it is a male?
What is the probability of female given that they were right-handed?
What is the probability of being left-handed?
P(LH | M) = 12/60 = 0.20
P(F| RH) = 42/90 = 0.467
P(LH) = 20/110 = 0.182

9. Tree Diagram Right-handed Male Left-handed Sex Right-handed Female
0.8
Right-handed
0.44
Male
0.55
0.2
Left-handed
0.11
Sex
Right-handed
0.84
0.378
0.45
Female
Left-handed
0.16
0.072

10. 3 possible outcomes  (1- P(a)- P(b)) = 1 – 0.24 – 0.24 = 0.52
Example 1
A construction firm has bid on two different contracts. Let B1 be the event that the first bid is successful and B2, that the second bid is successful. Suppose that P(B1) = .4, P(B2) = .6 and that the bids are independent. What is the probability that: a) both bids are successful? b) neither bid is successful? c) is successful in at least one of the bids?
Independent  P(B1) • P(B2) = 0.4 • 0.6 = 0.24
Independent  (1- P(B1)) • (1 - P(B2)) = 0.6 • 0.4 = 0.24
3 possible outcomes  (1- P(a)- P(b)) = 1 – 0.24 – 0.24 = 0.52 or
P(B1) • (1 – P(B2)) + (1 – P(B1)) • P(B2) = 0.4 • • 0.6 = 0.52

11. Example 2
Given that P(A) = .3 , P(B) = .6, and P(B|A) = .4 find: a) P(A and B) b) P(A or B) c) P(A|B)
P(A and B)
P(B|A) = so P(A and B) = P(B|A)•P(A)
P(A)
P(A and B) = 0.4 • 0.3 = 0.12
P(A or B) = P(A) + P(B) – P(A and B) = – 0.12
= 0.58
P(A and B)
P(A|B) = = = 0.2
P(B)

12. Example 3
Given P(A | B) = 0.55 and P(A or B) = 0.64 and P(B) = 0.3. Find P(A).
P(A and B)
P(A|B) = so P(A and B) = P(A|B)•P(B)
P(B)
P(A and B) = 0.55 • 0.3 = 0.165
P(A or B) = P(A) + P(B) – P(A and B)
P(A) = P(A or B) – P(B) + P(A and B)
= – =

13. Example 4
If 60% of a department store’s customers are female and 75% of the female customers have a store charge card, what is the probability that a customer selected at random is female and had a store charge card?
Let A = female customer and let B = customer has a store charge card
P(A and B)
P(B|A) = so P(A and B) = P(B|A)•P(A)
P(A)
P(A and B) = 0.75 • 0.6 = 0.45

14. Example 5
Suppose 5% of a box of 100 light blubs are defective. If a store owner tests two light bulbs from the shipment and will accept the shipment only if both work. What is the probability that the owner rejects the shipment?
P(reject) = P(at least one failure)
= 1 – P(no failures)
= 1 – P(1st not defective) • P(2nd not defective | 1st not defective)
= 1 – (95/100) • (94/99(
= 1 –
= or 9.8% of the time

15. Question to Ponder Dan can hit the bulls eye ½ of the time
Daren can hit the bulls eye ⅓ of the time
Duane can hit the bulls eye ¼ of the time
Given that someone hits the bulls eye, what is the probability that it is Dan?
Out of 36 throws, 13 hit the target. Dan had 6 of them, so P(Dan | bulls eye) = 6/13
= 0.462
Hit
Miss
Dan (p=1/2) 6
Daren (p=1/3) 4 8
Duane (p=1/4) 3 9

16. Summary and Homework
Summary
Homework
Day Two: 6.72, 73, 76, 81, 82