Download P(A B)

Lesson 6 – 3b
General Probability Rules
Taken from http://www.pendragoncove.info/statistics/statistics.htm
And modified slightly
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.
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
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
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
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
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) = 1/6
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
Contingency Tables
Right handed
Left handed
Total
Male
48
Female
42
Total
90
12
60
8
50
20
110
1. What is the probability of left-handed given that
it is a male? P(LH | M) = 12/60 = 0.20
2. What is the probability of female given that they
were right-handed?
P(F| RH) = 42/90 = 0.467
3. What is the probability of being left-handed?
P(LH) = 20/110 = 0.182
Tree Diagram
0.8
Right-handed
0.44
0.2
Left-handed
0.11
0.84
Right-handed
0.378
0.16
Left-handed
0.072
Male
0.55
Sex
0.45
Female
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?
Independent  P(B1) • P(B2) = 0.4 • 0.6 = 0.24
b) neither bid is successful?
Independent  (1- P(B1)) • (1 - P(B2)) = 0.6 • 0.4 = 0.24
c) is successful in at least one of the bids?
1 – P(neither bid is successful) = 1 - .24 = .76
Example 2
Given that P(A) = .3 , P(B) = .6, and P(B|A) = .4 find:
a) P(A and 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
b) P(A or B)
P(A or B) = P(A) + P(B) – P(A and B)
= 0.3 + 0.6 – 0.12
= 0.78
c) P(A|B)
P(A and B)
0.12
P(A|B) = ----------------- = -------- = 0.2
P(B)
0.6
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)
Rearrange Formula
P(A) = P(A or B) – P(B) + P(A and B)
= 0.64 – 0.3 + 0.165
= 0.505
Example 4
If 60% of a department store’s customers are female
and 75% of the female customers have a store charge
card (assuming these events are independent), 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(A)•P(B)
P(A and B) = 0.75 • 0.6 = 0.45
Example 5 - Independent
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?
(Assume each light bulb is independent)
P(accept) = P(1st not defective AND 2nd not defective)
=
.95 • .95
=
.9025
P(reject) = 1 - P(accept)
= 1 – .9025
= .0975 or 9.75% of the time
Example 5 - Dependent
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?
(Assume each light bulb is dependent)
P(1st not defective) = 95/100
P(2nd not defective | 1st not defective) = 94/99
P(accept) = P( both are not defective)
= ( 95/100) • (94/99)
= .9020
P(reject) = 1 - P(accept)
= 1 – .9020
= .098 or 9.8% of the time
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?
Hit Miss
Dan (p=1/2)
6
6
Daren (p=1/3)
4
8
Duane (p=1/4)
3
9
Out of 36 throws, 13 hit the
target. Dan had 6 of them,
so
P(Dan | bulls eye) = 6/13
= 0.462
Summary and Homework
• Summary
• Homework
– Day Two: 6.72, 73, 76, 81, 82