slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Let’s consider this problem… PowerPoint Presentation
Download Presentation
Let’s consider this problem…

Loading in 2 Seconds...

play fullscreen
1 / 29

Let’s consider this problem… - PowerPoint PPT Presentation


  • 84 Views
  • Uploaded on

Let’s consider this problem…. The table below gives the proportion of time that the gerbil spends in each compartment. Compartment. A. B. C. D. Proportion. 0.25. 0.20. 0.35. 0.30. 9.(a) Is this a valid probability function? Explain. No. 0.25 + 0.20 + 0.35 + 0.30 = 1.1,

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Let’s consider this problem…' - norman


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

Let’s consider this problem…

The table below gives the proportion of time that the gerbil

spends in each compartment.

Compartment

A

B

C

D

Proportion

0.25

0.20

0.35

0.30

9.(a) Is this a valid probability function? Explain.

No. 0.25 + 0.20 + 0.35 + 0.30 = 1.1,

which does not equal 1.

9.(b) Is there a problem with Alrik’s reasoning? Explain

Yes. Since the gerbil must always be

in exactly one of the four rooms, the

proportions must add up to 1.

slide2

Let’s consider this problem…

The table below gives the proportion of time that the gerbil

spends in each compartment.

Compartment

A

B

C

D

Proportion

0.40

0.30

0.20

0.10

10. Suppose Alrik determines that his gerbil spends time in the

four compartments A, B, C, and D in the ratio 4:3:2:1. What

proportions should he fill in the table above? Is this a valid

probability function?

Since 4 + 3 + 2 + 1 = 10, we can divide each

number in the ratio by 10 and get the

proportions relative to the whole.

slide4

Some Useful Diagrams for Probability

Venn Diagram – good for visualizing

relationships between events in a given

sample space

Tree Diagram – good for visualizing

problems that rely on the Multiplication

Principle of Probability.

slide5

In a large high school, 54% of the students are girls and 62%

of the students play sports. Half of the girls at the school play

sports.

(a) What percentage of the students who play sports are boys?

(b) If a student is chosen at random, what is the probability that

it is a boy who does not play sports?

(a) Ratio of boys who play sports

to all students who play sports:

0.35

0.27

0.35

0.27

= 0.5645

0.62

Girls

Sports

0.11

  • About 56.45% of the

students who play sports

are boys

Venn Diagram

slide6

In a large high school, 54% of the students are girls and 62%

of the students play sports. Half of the girls at the school play

sports.

(a) What percentage of the students who play sports are boys?

(b) If a student is chosen at random, what is the probability that

it is a boy who does not play sports?

(b) A boy that also does not play

sports is represented outside

of both circles:

0.27

0.35

0.27

0.11

Girls

Sports

  • There is an 11% chance

that a student chosen at

random is a boy who does

not play sports

0.11

Venn Diagram

slide7

Two identical cookie jars are on a counter. Jar A contains 2

chocolate chip and 2 peanut butter cookies, while jar B contains

1 chocolate chip cookie. We select a cookie at random. What

is the probability that it is a chocolate chip cookie?

There are 5 total cookies, 3 of which are chocolate chip…

P(chocolate chip) = 3/5, right???

No  the fact that the cookies are in different jars means that

they are not equally likely outcomes… (i.e., the chocolate chip

cookie in jar B is more likely to be chosen than any single

cookie in jar A)

We need to visualize this experiment as a two-step process:

First choose a jar, then choose a cookie from that jar!!!

slide8

Two identical cookie jars are on a counter. Jar A contains 2

chocolate chip and 2 peanut butter cookies, while jar B contains

1 chocolate chip cookie. We select a cookie at random. What

is the probability that it is a chocolate chip cookie?

CC

0.125

The event “chocolate chip”

is a set that contains three

outcomes. Add their

probabilities:

0.25

0.25

CC

0.125

Jar

A

0.25

0.5

PB

0.125

0.25

P(chocolate chip) =

PB

0.125

0.125 + 0.125 + 0.5

0.5

1

= 0.75

Jar

B

CC

0.5

Tree Diagram

slide9

Conditional Probability

Our previous example is one of conditional probability, since

the “cookie” outcome is dependent on the “jar” outcome.

Notation:

P(A|B)

  • Read “P of A given B”
  • Meaning “the probability of the event A, given
  • that event B occurs”

Previous Example:

2

P(chocolate chip | jar A) =

P(chocolate chip | jar B) = 1

4

Multiplication Principle of Probability:

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

slide10

Conditional Probability

Conditional Probability Formula:

If the event B depends on the event A, then

P(A and B)

P (B | A) =

P(A)

slide11

Conditional Probability

Suppose we have drawn a cookie at random from one

of the jars described in the previous example. Given

that it is chocolate chip, what is the probability that it

came from jar A?

P(jar A | chocolate chip) =

P(jar A and chocolate chip)

(1/2)(2/4)

1

=

=

P(chocolate chip)

0.75

3

slide12

Guided Practice

A and B are events in a sample space S such that P(A) = 0.7,

P(B) = 0.4, and P(A and B) = 0.2.

1. Find the probability that A occurs but B does not.

P = 0.5

2. Find the probability that B occurs but A does not.

P = 0.2

3. Find the probability that neither A nor B occurs.

P = 0.1

0.5

0.2

0.2

A

B

0.1

slide13

Guided Practice

A and B are events in a sample space S such that P(A) = 0.7,

P(B) = 0.4, and P(A and B) = 0.2.

4. Are events A and B independent? (That is, does

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

NO

P(A and B)

0.2

P(A | B) =

=

= 0.5 = P(A)

P(B)

0.4

0.5

0.2

0.2

A

B

0.1

slide14

Guided Practice

If the school cafeteria serves meat loaf, there is a 70% chance

that they will serve peas. If they do not serve meat loaf, there is

a 30% chance that they will serve peas anyway. The students

know that meat loaf will be served exactly once during the 5-day

week, but they do not know which day. If tomorrow is Monday,

what is the probability that

(a) the cafeteria serves meat loaf?

P = 0.2

(b) the cafeteria serves meat loaf and peas?

P = 0.14

(c) the cafeteria serves peas?

P = 0.38

0.7

ML

P

0.14

0.2

NP

0.06

0.3

NML

0.3

P

0.24

0.8

0.7

NP

0.56

let s start with a familiar experiment
Let’s start with a familiar experiment

You roll a fair die four times. Find the probability that you roll:

(a) all 3’s.

There is a probability 1/6 of rolling a three each time…

Multiplication Principle:

4

P(rolling 3 four times) = (1/6)

let s start with a familiar experiment1
Let’s start with a familiar experiment

You roll a fair die four times. Find the probability that you roll:

(b) no 3’s.

There is a probability 5/6 of rolling a non-three each time…

Multiplication Principle:

4

P(rolling non-3 four times) = (5/6)

let s start with a familiar experiment2
Let’s start with a familiar experiment

You roll a fair die four times. Find the probability that you roll:

(c) exactly two 3’s.

Probability of rolling two 3’s followed by two non-3’s:

2

2

(1/6) (5/6)

However, there are other outcomes to consider!!!

The two 3’s could occur anywhere among the four rolls

 In how many ways?

ways

let s start with a familiar experiment3
Let’s start with a familiar experiment

You roll a fair die four times. Find the probability that you roll:

(c) exactly two 3’s.

So we have 6 possible outcomes, each with probability:

2

2

(1/6) (5/6)

P(exactly two 3’s) =

let s start with a familiar experiment4
Let’s start with a familiar experiment

The forms of these three answers should look familiar…

Let’s let p = 1/6 and q = 5/6:

4

P(four 3’s) = p

These are three terms in the

expansion of (p + q) !!!

4

4

P(no 3’s) = q

4

2

2

P(two 3’s) = p q

2

let s start with a familiar experiment5
Let’s start with a familiar experiment

The forms of these three answers should look familiar…

Let’s let p = 1/6 and q = 5/6:

The terms in the expansion

give the exact probabilities of 4, 3, 2, 1, and 0 threes (respectively)

when tossing a fair die four times!!!

For this reason, such an example yields what is called a

Binomial Probability Distribution

slide22

Theorem: Binomial Distribution

Suppose an experiment consists of n independent repetitions of

an experiment with two outcomes, called “success” and “failure.”

Let P(success) = p and P(failure) = q. (Note that q = 1 – p.)

n

Then the terms in the binomial expansion of (p + q) give the

respective probabilities of exactly n, n – 1,…, 2, 1, 0 successes.

The distribution is shown on the following slide…

slide23

Number of successes out of

n independent repetitions

Probability

more practice problems
More Practice Problems

Suppose Michael makes 90% of his free throws. If he shoots

20 free throws, and if his chance of making each one is

independent of the other shots, what is the probability that he

makes (a) all 20, (b) exactly 18, and (c) at least 18?

This situation yields a binomial probability distribution,

with p = success = 0.9, and q = failure = 0.1.

(a) P(20 successes) =

(b) P(18 successes) =

more practice problems1
More Practice Problems

Suppose Michael makes 90% of his free throws. If he shoots

20 free throws, and if his chance of making each one is

independent of the other shots, what is the probability that he

makes (a) all 20, (b) exactly 18, and (c) at least 18?

(c) P(at least 18 successes) =

P(18) + P(19) + P(20)

more practice problems2
More Practice Problems

A new medical test for a deadly virus is 0.7% likely to produce a

false positive result (i.e., the test indicates the presence of the

virus when it is not actually present). If 60 individuals are tested,

what is the probability that (a) there will be 3 false positives,

and (b) at least one false positive?

Here, p = 0.007 and q = 0.993

(a) P(3 false) =

more practice problems3
More Practice Problems

A new medical test for a deadly virus is 0.7% likely to produce a

false positive result (i.e., the test indicates the presence of the

virus when it is not actually present). If 60 individuals are tested,

what is the probability that (a) there will be 3 false positives,

and (b) at least one false positive?

Here, p = 0.007 and q = 0.993

(b) P(at least 1 false) = 1 – P(no false)

more practice problems4
More Practice Problems

Ten dimes, dated 1990 through 1999 are tossed. Find the

probability of each of the following events.

1. Heads on the 1990 dime only.

P(H-T-T-T-T-T-T-T-T-T)

2. Heads on the 1991 and 1996 dimes only.

P(T-H-T-T-T-T-H-T-T-T)

more practice problems5
More Practice Problems

Ten dimes, dated 1990 through 1999 are tossed. Find the

probability of each of the following events.

3. Heads on all but one dime.

P(9H and 1T)

4. Exactly two heads.

P(2H and 8T)