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## PowerPoint Slideshow about '2.1 Conditional Probability and Multiplication Rule' - mike_john

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### 2.1 Conditional Probability and Multiplication Rule

### 2.1 Conditional Probability and Multiplication Rule

### 2.1 Conditional Probability and Multiplication Rule

### 2.1 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.2 Conditional Probability and Multiplication Rule

### 2.3 Independence

### 2.3 Independence

### 2.3 Independence

### 2.3 Independence

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

### 2.4 A technique for Finding P(A or B or C or …)

Sometimes we are concerned with probabilities about some portion of the sample space.

Example 1: The probability that a person has an annual income over $100, 000 would be different than the probability that a college graduate has an annual income over $100, 000. In this example, we are reducing the sample space. The reduced sample space consists of college graduates.

Example 2: One chip is selected at random from a box containing five chips numbered 1, 2, 3, 4, 5. So S={1, 2, 3, 4, 5} is an equally probable sample space.

What is the probability of selecting a 1?

Solution: Let B={1}, then P(A)=1/5.

(b) Suppose we are told that the outcome is an odd number. Let A={1, 2, 3}. We are given that the outcome is in A. What is the probability of getting a 1?

Solution: The answer is 1/3. We write this as P(B| A)=1/3. The vertical line separating events B and A means “given” and P(B|A) is called the conditional probability of B given A.

Another method for finding a conditional probability is by using the formula:

To solve Example 2 using this formula, we first find the event “B and A.” But the event “B and A”={1} since “B and A” includes outcomes common to both event B and event A. Relative to the original sample space S,

P(B and A)=P({1})=1/5, and P(A)=P({1, 2, 3})=3/5.The formula gives

2.2 Multiplication Rule

This is called the multiplication rule, it gives us a method for finding P(B and A) when the conditional probability is known.

The multiplication rule also holds when we have more than two events. For three events, the multiplication is

Example 3: Urns I, II, III each contain four chips numbered 1, 2, 3, 4. One chip is selected at random from each earn. Find the probability of getting three different numbers.

Solution: Let A=“all the three numbers drawn are different”;

B=“any number is drawn from I”;

C=“number from II is different than number from I”;

D= =“number from III is different than numbers from I and II.”

Example 4: (Birthday Problem.) Suppose there are 65 students in a room(no twins). What is the probability that all 65 have different birthdays? Assume all 365 birthdays are equally likely.

Solution: Imagine 65 boxes, one for each student. Each box contains 365 chips which correspond to 365 days of the year. We think in terms of selecting one chip from each box corresponds to selecting a birthday, at random, for each student

Let A=“all 65 students have different birthdays”;

B=“any birthday is selected for student 1”;

C=“birthday for student 2 is different than birthday for student 1”;

D=“birthday for student 3 is different than birthdays for student 1 and 2,” etc.

Then A=B and C and D and…;

P(A)=P(B and C and D…);

Example 5: In the above example, find the probability of at least two students in the room have the same birthday.

Solution: Not A=“At least two students in the room have the same birthday.” Therefore, P(Not A)=1-P(A)=1-.01=.99

Remark: Using the same type of reasoning, it is possible to show that we need only 23 students in a room to have better chance a 50-50 chance that at least two students have the same birthday.

Example 6: Johnny Carson, on hearing about the birthday problem, once observed during the Tonight Show that there were about 120 people in his audience. He asked all audience members who shared his birthday of October 23 to raise their hands. To Johnny’s surprise, there were no raised hands. Johnny’s mistake was that while the probability of at least two audience members having the same birthday is large, the probability that at least one audience member has a particular birthday which matches his is quite small.

To see this, let

Let A=“No audience member has an October 23 birthday,”;

B=“first audience member does not have an October 23 birthday”;

C=“second audience member does not have an October 23 birthday”;

D=“=“third audience member does not have an October 23 birthday,” etc.

Then A=B and C and D and…;

P(A)=P(B and C and D…);

Example 7: During the 120-day period between November 1968 and February 1969, there were 22 commercial hijacked to Cuba. On a day when there were two hijacking, the New York Times regarded the occurrence of more than one hijacking on the same day as a “sensational” and improbable coincidence.”

Model: Suppose 22 balls are tossed into 120 boxes at random. Label the 120 boxes: November 1, 1968,

November 2, 1968,…, February 28, 1969. When a ball lands in a box it corresponds to a hijacking on that day

The event A, B are said to be independent events if the occurrence or non-occurrence of event A does not affect the probability of the occurrence of event B. That is,

P(B|A)=P(B|not A)=P(B).

Either of two equations can be used to check for independence:

(i) P(B|A)=P(B) (equivalently

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

(ii) P(A and B)=P(A)P(B).

Example 8:

A box contains four chips numbered 1, 2, 3, 4. Two chips are drawn, at random, without replacement from the box. Let A =“sum of the numbers drawn is even” and B=“one of the numbers drawn is 4.” Are A and B independent?

Example 9:

Select a card from an ordinary deck of 52 cards. Let A=“ace” and B=“spade.” Are A and B independent?

Example 10:

A fair coin is tossed three times. Find P(at least one head).

Solution: P(at least one head)

=1-P( 3 tails)

=1- P(T)P(T)P(T)

=1-(1/2)(1/2)(1/2)=7/8

In general: If A, B, C, …are independent, then

P(A or B or C or …)=1- P(“not A” and “not B” and “not C” …)

=1-[1-P(A)][1-P(B)][1-P(C)]……

Class Exercise:

A fair coin is tossed 40 times. Find P(at least one head).

Example 11: In 1978 Pete Rose set a National League record by hitting safely in each of 44 consecutive games. His life time batting average was .303. Also, assume he came to bat four times each game and his chances of getting a hit on each at bat did not depend on previous at bats. Find the probability that

He got at least one hit in a given game.

He got at least one hit in each of 44 consecutive games.

Solution:

P( at least one hit in a game)

=1-P(no hits in a game)

=1-P(“out on 1st at bat” and… “out on 4th at bat” )

=1-P(“out on 1st an bat”)…P( “out on 4th an bat” )

=1-(.697)(.697)(.697)(.697)=.764

Solution:

(ii) P( at least one hit in each of 44 consecutive games)

=P( at least one hit in game 1 and at least one hit in game 2 and…)

= P( at least one hit in game 1)P(at least one hit in game 2)…

=

Example 12: Three sisters who live near Provo, Utah all gave birth on March 11, 1998. This is obviously a rare event. How rare: This raises the question, “what is the probability that three sisters will give birth on the same day?”

Solution: We focus on three possible interpretations of the questions:

If each sisters will give birth in a given year, what is the probability that all of them will give birth on March 11?

P(all the three sisters will give birth on March 11)=

2. If each of three sisters will give birth in a given year, what is the probability that all three will give birth on the same day?

P(1st sister give birth on any day, and 2nd sister give birth on the same day as 1st sister, and 2nd sister give birth on the same day as 1st sister )=

What is the probability that somewhere in the United states, there are three sisters who will give birth on the same day sometime in a given year?

Suppose there are a total of 3 groups of sisters in the united states who will give birth in a given year.

P(One group of three sisters will give birth on the same day)=

P(One group of three sisters will not give birth on the same day)=

P(non of the three groups of three sisters will give birth on the same day)=

P(at least one group of three sisters will give birth on the same day)=

2.5 Problems

Homework / Class Exercises (Section 2.5)

Do problems # 1-6, 9-13, 15-17, 19-20, 21, 24, 27, 35

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