Objectives (BPS chapter 12)

General rules of probability

Independence and the multiplication rule

The general addition rule

Conditional probability

The general multiplication rule

Independence

Tree diagrams

In this example, the trials are independent only when you put

the coin back (“sampling with replacement”) each time.

Two events are independent if the probability that one event occurs

on any given trial of an experiment is not influenced in any way by

the occurrence of the other event.

Imagine coins spread out so that half were heads up, and half were tails up. Pick

a coin at random. The probability that it is heads-up is 0.5. But, if you don’t put it

back, the probability of picking up another heads-up coin is now less than 0.5.

Without replacement, successive trials are not independent.

Independent versus disjoint events

The multiplication rule for independent events

Two events A and B are independent if knowing

that one occurs does not change the probability

that the other occurs. If A and B are

independent, P(A and B) = P(A) P(B)

If A is the event that you roll a “1” on a red 6-sided die, and B is the event that you

roll a “1” on a blue 4-sided die, what is the probability that if you roll both at the

same time, you will roll a “1” on both dice? Since the two dice are independent,

P(both “1”s) = P(“1” on red die) P(“1” on blue die)

= (1/6) (1/4) = 1/24

The general addition rule

General addition rule for any two events A and B:

The probability that A occurs,

or B occurs, or both events occur is:

P(A or B) = P(A) + P(B) – P(A and B)

What is the probability of randomly drawing either an ace or a heart from a pack of

52 playing cards? There are 4 aces in the pack and 13 hearts. However, one card

is both an ace and a heart. Thus:

P(ace or heart) = P(ace) + P(heart) – P(ace and heart)

= 4/52 + 13/52 - 1/52 = 16/52 ≈ 0.3

Conditional probability

Conditional probabilities reflect how the probability of an event can

change if we know that some other event has occurred/is occurring.

Example: The probability that a cloudy day will result in rain is different if

you live in Los Angeles than if you live in Seattle.

Our brains effortlessly calculate conditional probabilities, updating our

“degree of belief” with each new piece of evidence.

The conditional probability

of event B given event A is:

(provided that P(A) ≠ 0) )(

)()|(

AP

BandAPABP

If A and B are independent, P(B | A) =P(B).

The general multiplication rule The probability that any two events, A and B, both occur is:

P(A and B) = P(A)P(B|A)

This is the general multiplication rule.

If A and B are independent, then P(A and B) = P(A)P(B) A and B are independent when they have no influence on each other’s occurrence.

Another way of showing this is P (B | A) = P (B) or P (A | B) = P (A).

What is the probability of randomly drawing either an ace or heart from a pack of

52 playing cards? There are four aces in the pack and thirteen hearts.

P(heart|ace) = 1/4 P(ace) = 4/52

P(ace and heart) = P(ace)* P(heart|ace) = (4/52)*(1/4) = 1/52

Notice that heart and ace are independent events.

P(ace or heart) =P(ace)+P(heart)-P(ace and heart)=4/52+13/52-1/52=16/52

Probability trees

Conditional probabilities can get complex and it is often a good strategy

to build a probability tree that represents all possible outcomes

graphically and assigns conditional probabilities to subsets of events.

0.47Internet user

Tree diagram for chat room

habits for three adult age

groups.

P(chatting) = 0.136 + 0.099 + 0.017

= 0.252

About 25% of all adult Internet users visit chat rooms.

Breast cancer screening

Cancer

No cancer

Mammography

Positive

Negative

Positive

Negative

Disease incidence

Diagnosis sensitivity

Diagnosis specificity

False negative

False positive

0.0004

0.9996

0.8

0.2

0.1

0.9Incidence of breast cancer among women ages 20–30 Mammography

performance

If a woman in her 20s gets screened for breast cancer and receives a positive

test result, what is the probability that she has breast cancer?

She could either have a positive test and have breast cancer, or have a positive

test but not have cancer (false positive).

Cancer

No cancer

Mammography

Positive

Negative

Positive

Negative

Disease incidence

Diagnosis sensitivity

Diagnosis specificity

False negative

False positive

( )( | )

( ) ( )

0.0004*0.80.3%

0.0004*0.8 0.9996*0.1

P cancer and posP cancer pos

P cancer and pos P nocancer and pos

0.0004

0.9996

0.8

0.2

0.1

0.9Incidence of breast cancer among women ages 20–30 Mammography

performance

Possible outcomes given the positive diagnosis: positive test and breast cancer,

or positive test but not cancer (false positive).

This value is called the positive predictive value, or PV+. It is an important piece

of information but, unfortunately, is rarely communicated to patients.

Musical styles other than rock and pop are becoming more popular. A survey of college students finds that 40% like country music, 30% like gospel music, and 10% like both country and gospel music.

What is the probability that a randomly selected college student likes gospel but not country music?

Practice Problem 1

A: likes country music, P(A)=0.4 B: likes gospel music, P(B)=0.3 A and B: likes both, P(A and B)=0.1

2.0)( BAP c

What is the probability that a randomly selected college student doesn’t like gospel or country music?

What is the probability that the student chosen likes gospel music given that he/she likes country music?

Practice Problem 1

4.06.01)(1)( BAPBAP cc

25.04.0

1.0

)(

)()|(

AP

BAPABP

You and a friend decide to have lunch at a nearby Chinese restaurant. The table below lists the six meals you and your friend ever order, along with the price of the dish, the probabilities that you will order each dish, and the probabilities your friend will order each dish. Assume that your choice of a meal has no bearing on your friend’s choice (and vice versa).

Practice Problem 2

Luncheon special Price Your probability

Your friend’s probability

Roast Pork with Bean Sprouts

$4.00 0.05 0.20

Szechuen Chicken (spicy)

$4.50 0.10 0.10

Curry Shrimp (spicy)

$4.75 0.30 0.30

Beef with Broccoli $4.50 0.10 0.10

Chicken Wings $3.25 0.10 0.10

Delight of Three $5.00 0.35 0.20

1. Let A be the event that you order a spicy lunch. Find P(A).

1. P(A)= P (Szechuen Chicken or Curry Shrimp) =0.1+0.3=0.4�

2. Let B be the event you do not order a spicy lunch, and C be the event you order a lunch that costs less than $4.60. Find P(B and C).

P (B and C) = P(Roast Pork with Bean Sprouts or Beef with Broccoli or Chicken Wings)

= 0.05 + 0.1+ 0.1=0.25

3. With B and C defined in the previous part, find P(B or C).

P(B or C) = P( ) + P(C) − P(B and C) = 0.6 + 0.35 − 0.25 = 0.7𝐵4. What is the probability that both you and your friend order spicy lunches?

P(A and E) P(A)P(E) = 0.4 × 0.4 = 0.16

Practice Problem 2