Chapter 6: Probability

The study of randomness

Ch 6.2 Probability Models

• Proportion of heads to tails in a few tosses will be erratic but after thousands of tosses will approach the expected .5 probability

Proportion of heads to tails in a few tosses will be erratic but after

thousands of tosses will approach the expected .5 probability

Probability models have two parts:

A list of possible outcomes

A probability for each outcome.

Sample Space

• To specify S we must state what constitutes an individual outcome, then which outcomes can occur (can be simple or complex)

– Ex: coin tossing, S = {H, T}

– Ex: US Census: If we draw a random sample of 50,000 US households, as the survey does, the S contains all 50,000

Rolling two dice At a casino- 36 possible outcomes when we

roll 2 dice and record the up-faces in order (first die, second die)

Gamblers care only about number of dots face up so the sample space for that is: S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Techniques for finding outcomes

• 1. Tree diagram

• For tossing a coin then rolling a die

• 2. Multiplication Principle – 2x6 = 12 for same example

• 3. Organized list:

– H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6

With/without replacement

• If you take a card from a deck of 52, don’t put it back, then draw your 2nd card etc., that’s without replacement.

– Ex: how many different 3 digit numbers can you make: 10x9x8 = 720

• If you take a card, write it down, put it back, draw 2nd card etc., that’s with replacement.

– Ex: 10x10x10 = 1000

Probability Rules, pg 342

• 1. Any probability is a number between 0 and 1

• 2. The sum of the probabilities of all possible outcomes = 1

• 3. If two events A and B are disjoint (mutually exclusive, no outcomes in common), then P(A or B) = P(A) + P(B)

• 4. The probability that an event doesn’t occur is 1 minus the probability that it does occur

– P(Ac) = 1 – P(A)

Venn diagrams help!

Probabilities in a Finite Space

Ex: Probability of rolling a 5?

P(roll a 5 with 2 die) =

P(1,4) + P(3,2) + P(2,3) + P(4,1)

= 1/36 + 1/36 + 1/36 + 1/36

= 1/9 or .111

Independence & the Multiplication Rule

• To find the probability for BOTH events A and B occurring

– Example: Suppose you plan to toss a coin twice, and want to find the probability of rolling a head on both tosses.

– A = first toss is a head, B = second toss is a head. So (1/2)(1/2) = ¼. We expect to flip 2 heads on 25% of all trials. The more times we repeat this, the closer our average probability will get to 25%.

• The multiplication rule applies only to independent events; can’t use it if events are not independent!

Independent or not? Coin toss I: Coin has no memory and coin tossers cannot

influence fall of coin

Drawing from deck of cards NI: First pick, probability of red is 26/52 or .5.

Once we see the first card is red, the probability of a red card in the 2nd pick is now 25/51 = .49

Taking an IQ test twice in succession NI

More applications of Probability Rules

• If two events A and B are independent, then their complements are also independent.

– Ex: 75% of voters in a district are Republicans. If an interviewer chooses 2 voters at random, the probability that the first is a Republican and the 2nd is not a republican is .75 x .25 = .1875

6.3 General Probability Rules

Addition Rule for Disjoint events

General Addition rule for Unions of 2 events

Example:

Deb and Matt are waiting anxiously to hear if they’ve been promoted. Deb guesses her probability of getting promoted is .7 and Matt’s is .5, and both of them being promoted is .3. The probability that at least one is promoted = .7 + .5 - .3 which is .9. The probability neither is promoted is .1.

The simultaneous occurrence of 2 events (called a joint event, such as deb and matt getting promoted) is called a joint probability.

Conditional Probability

The probability that we assign to an event can change if we know some other event has occurred. P(A|B): Probability that event A will happen under

the condition that event B has occurred.

Ex: Probability of drawing an ace is 4/52 or 1/13. If your are dealt 4 cards and one of them is an ace, probability of getting an ace on the 5th card dealt is 3/48 or 1/16 (conditional probability- getting an Ace given that one was dealt in the first 4).

In words, this says that for both of 2 events

to occur, first one must occur, and then,

given that the first event has occurred, the

second must occur.

Remember: B is the event whose probability we are

computing and A represents the info we are given.

6.3 Need to Know summary(print)

• Complement of an event A contains all outcomes not in A

• Union (A U B) of events A and B = all outcomes in A, in B, or in both A and B

• Intersection(A^B) contains all outcomes that are in both A and B, but not in A alone or B alone.

• General Addition Rule: P(AUB) = P(A) + P(B) – P(A^B)

• Multiplication Rule: P(A^B) = P(A)P(B|A)

• Conditional Probability P(B|A) of an event B, given that event A has occurred: P(B|A) = P(A^B)/P(A) when P(A) > 0

• If A and B are disjoint (mutually exclusive) then P(A^B) = 0 and P(AUB) = P(A) + P(B)

• A and B are independent when P(B|A) = P(B)

• Venn diagram or tree diagrams useful for organization.

Extended Multiplication rules

• The union of a collection of events is the event that ANY of them occur

• The Intersection of any collection of events is the event that ALL of them occur

Example • Only 5% of male high school basketball, baseball, and football players go on to

play at the college level. Of these only 1.7% enter major league professional sports. About 40% of the athletes who compete in college and then reach the pros have a career of more than 3 years. Define these events: – A = competes in college B = competes pro C = pro career longer than 3 years

– P(A) = .05

– P(B|A) = .017

– P(C|A and B) = .400

– What is the probability a HS athlete will have a pro career more than 3 years? The probability we want is therefore

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

• = .05 x .017 x .40 = .00034

– So, only 3 of every 10,000 high school athletes can expect to compete in college and have a pro career of more than 3 years.

Extended tree diagram + chat room example

• 47% of 18 to 29 age chat online, 21% of 30 to 49 and 7% of 50+

• Also, need to know that 29% of all internet users are 18-29 (event A1), 47% are 30 to 49 (A2) and the remaining 24% are 50 and over (A3). – What is the probability that a randomly chosen user of the

internet participates in chat rooms (event C)? – Tree diagram- probability written on each segment is the

conditional probability of an internet user following that segment, given that he or she has reached the node from which it branches.

– (final outcome is adding all the chatting probabilities which = .2518)

Bayes Rule

• Another question we might ask- what percent of adult chat room participants are age 18 to 29?

• P(A1|C) = P(A1 and C) / P(C) = .1363/.2518 = .5413 *since 29% of internet users are 18-29, knowing that someone chats increases the

probability that they are young! Formula sans tree diagram: P(C) = P(A1)P(C|A1) + P(A2)P(C|A2) + P(A3)P(C|A3)