Lesson 5 - 4

Conditional Probability and the General Multiplication Rule

Objectives• Compute conditional probabilities

• Compute probabilities using the General Multiplication Rule

Vocabulary• Conditional Probability – probability that an event F

occurs, given that event E has occurred.

P(E|F) is the symbols for the above and is readprobability of F given E has occurred.

Conditional Probability Rule

If E and F are any two events, then

P(E and F) N(E and F)P(E|F) = ----------------- = ----------------

P(F) N(F)

N is the number of outcomes

General Multiplication Rule

The probability that two events E and F both occur is

P(E and F) = P(E) ∙ P(E|F)

Independence in Terms of Conditional Probability

Two events E and F are independent if P(E|F) = P(F)

Example: P(E=Rolling a six on a single die) = 1/6

P(F=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(E|F) = P(F) so E and F are independent

Contingency Tables

Male Female Total

Right handed 38 42 80

Left handed 12 8 20

Total 50 50 100

1. What is the probability of left-handed given that it is a male?

2. What is the probability of female given that they were right-handed?

3. What is the probability of being left-handed?

P(LH | M) = 12/50 = 0.24

P(F| RH) = 42/80 = 0.525

P(LH) = 20/100 = 0.20

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 = 0.52or

P(B1) • (1 – P(B2)) + (1 – P(B1)) • P(B2) = 0.4 • 0.4 + 0.6 • 0.6 = 0.52

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 and B) 0.12P(A|B) = ----------------- = -------- = 0.2 P(B) 0.6

P(A or B) = P(A) + P(B) – P(A and B) = 0.3 + 0.4 – 0.12 = 0.58

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)

= 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, 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

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 – 0.9020

= 0.098 or 9.8% of the time

Summary and Homework

• Summary– Conditional probabilities P(F|E) represent the

chance that F occurs, given that E occurs also– The General Multiplication Rule applies to “and”

problems for all events and involves conditional probabilities

• Homework– pg 289 - 292: 3, 5, 7, 9, 12, 19, 27, 31, 32