condition prob

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Practice with Conditional Probability Topic Index | Algebra Index | Regents Exam Prep Center Answer the following questions dealing with conditional probability. 1. You roll two dice. The first die shows a ONE and the other die rolls under the table and you cannot see it. Now, what is the probability that both die show ONE? Choose: 1 : 3 1 : 6 1 : 36 9 : 36

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Practice with

Conditional Probability

Topic Index | Algebra Index | Regents Exam Prep Center

Answer the following questions dealing with conditional probability.

1.

You roll two dice. The first die shows a ONE and the other die rolls under the table and you cannot see it. Now, what is the probability that both die show ONE?

Choose:

1 : 3

1 : 6

1 : 36

9 : 36

2. You decide to tell your fortune by drawing two cards from a standard deck of 52 cards. What is the probability of drawing two cards of the same suite in a row? The cards are not replaced in the deck.

Choose:

13/52

12/51

13/51

12/52

3.

A new bag of golf tees contains 10 red tees, 10 orange tees, 10 green tees and 10 blue tees. You empty the tees into your golf bag. What is the probability of grabbing out two tees of the same color in a row for you and your partner?

Choose:

1/40

10/40

9/39

9/40

4.

In a library box, there are 8 novels, 8 biographies, and 8 war history books. If Jack selects two books at random, what is the probability of selecting two different kinds of books in a row?

Choose:

16/23

16/24

8/24

7/23

5.

What is the probability that the sum of two die will be greater than 8, given that the first die is 6?

Choose:

1/2

3/4

2/3

7/12

6.

What is the probability of drawing two aces from a standard deck of cards, given that the first card is an ace?

The cards are not returned to the deck.

Choose:

1/663

1/221

4/52

3/51

7.

A new superman MasterCard has been issued to 2000 customers. Of these customers, 1500 hold a Visa card, 500 hold an American Express card and 40 hold a Visa card and an American Express card. Find the probability that a customer chosen at random holds a Visa card, given that the customer holds an American Express card.

Choose:

1/4

1/3

2/25

1/50

8.

A survey of middle school students asked: What is your favorite winter sport? The results are summarized below:

Favorite Winter Sport

Grade Snowboarding Skiing Ice Skating TOTAL

6th 68 41 46 155

7th 84 56 70 210

8th 59 74 47 180

TOTAL 211 171 163 545

Using these 545 students as the sample space, a student from this study is randomly selected.

a.) What is the probability of selecting a student whose favorite sport is skiing?

b.) What is the probability of selecting a 6th grade student?

c.) If the student selected is a 7th grade student, what is the probability that the student prefers ice-skating?

d.) If the student selected prefers snowboarding, what is the probability that the student is a 6th grade student?

e.) If the student selected is an 8th grade student, what is the probability that the student prefers skiing or ice-skating?

Topic Index | Algebra Index | Regents Exam Prep Center

Created by Donna Roberts

Copyright 1998-2012 http://regentsprep.org

Oswego City School District Regents Exam Prep Center

2

Conditional Probability Unit 6 > Lesson 9 of 12

Problem: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test? [IMAGE]

Analysis: This problem describes a conditional probability since it asks us to find the probability that the second test was passed given that the first test was passed. In the last lesson, the notation for conditional probability was used in the statement of Multiplication Rule 2.

Multiplication Rule 2: When two events, A and B, are dependent, the probability of both occurring is:

IMAGE

The formula for the Conditional Probability of an event can be derived from Multiplication Rule 2 as follows:

Start with Multiplication Rule 2.

Divide both sides of equation by P(A).

Cancel P(A)s on right-hand side of equation.

Commute the equation.

We have derived the formula for conditional probability.

Now we can use this formula to solve the problem at the top of the page.

Problem: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test? [IMAGE]

Solution:

P(Second|First) = P(First and Second) = 0.25 = 0.60 = 60%

P(First) 0.42

Let's look at some other problems in which we are asked to find a conditional probability.

Example 1: A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and then a white marble is 0.34, and the probability of selecting a black marble on the first draw is 0.47. What is the probability of selecting a white marble on the second draw, given that the first marble drawn was black?

Solution:

P(White|Black) = P(Black and White) = 0.34 = 0.72 = 72%

P(Black)0.47

Example 2: The probability that it is Friday and that a student is absent is 0.03. Since there are 5 school days in a week, the probability that it is Friday is 0.2. What is the probability that a student is absent given that today is Friday?

Solution:

P(Absent|Friday) = P(Friday and Absent) = 0.03 = 0.15 = 15%

P(Friday) 0.2

Example 3: At Kennedy Middle School, the probability that a student takes Technology and Spanish is 0.087. The probability that a student takes Technology is 0.68. What is the probability that a student takes Spanish given that the student is taking Technology?

Solution:

P(Spanish|Technology) = P(Technology and Spanish) = 0.087 = 0.13 = 13%

P(Technology) 0.68

Summary: The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A), read as the probability of B given A. The formula for conditional probability is:

The Venn Diagram below illustrates P(A), P(B), and P(A and B). What two sections would have to be divided to find P(B|A)? Answer

[IMAGE]

Exercises

Directions: Read each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX. If you make a mistake, choose a different button. Answer choices have been rounded to the nearest percent.

1. In New York State, 48% of all teenagers own a skateboard and 39% of all teenagers own a skateboard and roller blades. What is the probability that a teenager owns roller blades given that the teenager owns a skateboard?

87%

81%

123%

None of the above.

RESULTS BOX:

2. At a middle school, 18% of all students play football and basketball and 32% of all students play football. What is the probability that a student plays basketball given that the student plays football?

56%

178%

50%

None of the above.

RESULTS BOX:

3. In the United States, 56% of all children get an allowance and 41% of all children get an allowance and do household chores. What is the probability that a child does household chores given that the child gets an allowance?

137%

97%

73%

None of the above.

RESULTS BOX:

4. In Europe, 88% of all households have a television. 51% of all households have a television and a VCR. What is the probability that a household has a VCR given that it has a television?

173%

58%

42%

None of the above.

RESULTS BOX:

5. In New England, 84% of the houses have a garage and 65% of the houses have a garage and a back yard. What is the probability that a house has a backyard given that it has a garage?

77%

109%

19%

None of the above.

RESULTS BOX:

3

1.4.5 Solved Problems:

Conditional Probability

In die and coin problems, unless stated otherwise, it is assumed coins and dice are fair and repeated trials are independent.

Problem

You purchase a certain product. The manual states that the lifetime T of the product, defined as the amount of time (in years) the product works properly until it breaks down, satisfies

P(T≥t)=e−t5, for all t≥0.

For example, the probability that the product lasts more than (or equal to) 2 years is P(T≥2)=e−25=0.6703. I purchase the product and use it for two years without any problems. What is the probability that it breaks down in the third year?

Solution

Problem

You toss a fair coin three times:

What is the probability of three heads, HHH?

What is the probability that you observe exactly one heads?

Given that you have observed at least one heads, what is the probability that you observe at least two heads?

Solution

Problem

For three events A, B, and C, we know that

A and C are independent,

B and C are independent,

A and B are disjoint,

P(A∪C)=23,P(B∪C)=34,P(A∪B∪C)=1112 Find P(A),P(B), and P(C).

Solution

Problem

Let C1,C2,⋯,CM be a partition of the sample space S, and A and B be two events. Suppose we know that

A and B are conditionally independent given Ci, for all i∈{1,2,⋯,M};

B is independent of all Ci's.

Prove that A and B are independent.

Solution

Problem

In my town, it's rainy one third of the days. Given that it is rainy, there will be heavy traffic with probability 12, and given that it is not rainy, there will be heavy traffic with probability 14. If it's rainy and there is heavy traffic, I arrive late for work with probability 12. On the other hand, the probability of being late is reduced to 18 if it is not rainy and there is no heavy traffic. In other situations (rainy and no traffic, not rainy and traffic) the probability of being late is 0.25. You pick a random day.

What is the probability that it's not raining and there is heavy traffic and I am not late?

What is the probability that I am late?

Given that I arrived late at work, what is the probability that it rained that day?

Solution

Problem

A box contains three coins: two regular coins and one fake two-headed coin (P(H)=1),

You pick a coin at random and toss it. What is the probability that it lands heads up?

You pick a coin at random and toss it, and get heads. What is the probability that it is the two-headed coin?

Solution

Problem

Here is another variation of the family-with-two-children problem [1] [7]. A family has two children. We ask the father, "Do you have at least one daughter named Lilia?" He replies, "Yes!" What is the probability that both children are girls? In other words, we want to find the probability that both children are girls, given that the family has at least one daughter named Lilia. Here you can assume that if a child is a girl, her name will be Lilia with probability α≪1 independently from other children's names. If the child is a boy, his name will not be Lilia. Compare your result with the second part of Example 1.18.

Solution

Problem

If you are not yet confused, let's look at another family-with-two-children problem! I know that a family has two children. I see one of the children in the mall and notice that she is a girl. What is the probability that both children are girls? Again compare your result with the second part of Example 1.18. Note: Let's agree on what precisely the problem statement means. Here is a more precise statement of the problem: "A family has two children. We choose one of them at random and find out that she is a girl. What is the probability that both children are girls?"

Solution

Problem

Okay, another family-with-two-children problem. Just kidding! This problem has nothing to do with the two previous problems. I toss a coin repeatedly. The coin is unfair and P(H)=p. The game ends the first time that two consecutive heads (HH) or two consecutive tails (TT) are observed. I win if HH is observed and lose if TT is observed. For example if the outcome is HTHTT−−−, I lose. On the other hand, if the outcome is THTHTHH−−−−, I win. Find the probability that I win.

Solution

4

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Conditional Probability

How to handle Dependent Events

Life is full of random events! You need to get a "feel" for them to be a smart and successful person.

Independent Events

Events can be "Independent", meaning each event is not affected by any other events.

Example: Tossing a coin.

Each toss of a coin is a perfect isolated thing.

What it did in the past will not affect the current toss.

The chance is simply 1-in-2, or 50%, just like ANY toss of the coin.

So each toss is an Independent Event.

Dependent Events

But events can also be "dependent" ... which means they can be affected by previous events ...

Example: Marbles in a Bag

2 blue and 3 red marbles are in a bag.

What are the chances of getting a blue marble?

The chance is 2 in 5

But after taking one out the chances change!

So the next time:

if we got a red marble before, then the chance of a blue marble next is 2 in 4

if we got a blue marble before, then the chance of a blue marble next is 1 in 4

See how the chances change each time? Each event depends on what happened in the previous event, and is called dependent.

That is the kind of thing we look at here.

"Replacement"

Note: if we replace the marbles in the bag each time, then the chances do not change and the events are independent:

With Replacement: the events are Independent (the chances don't change)

Without Replacement: the events are Dependent (the chances change)

Tree Diagram

A Tree Diagram: is a wonderful way to picture what is going on, so let's build one for our marbles example.

There is a 2/5 chance of pulling out a Blue marble, and a 3/5 chance for Red:

We can even go one step further and see what happens when we select a second marble:

If a blue marble was selected first there is now a 1/4 chance of getting a blue marble and a 3/4 chance of getting a red marble.

If a red marble was selected first there is now a 2/4 chance of getting a blue marble and a 2/4 chance of getting a red marble.

Now we can answer questions like "What are the chances of drawing 2 blue marbles?"

Answer: it is a 2/5 chance followed by a 1/4 chance:

Did you see how we multiplied the chances? And got 1/10 as a result.

The chances of drawing 2 blue marbles is 1/10

Notation

We love notation in mathematics! It means we can then use the power of algebra to play around with the ideas. So here is the notation for probability:

P(A) means "Probability Of Event A"

In our marbles example Event A is "get a Blue Marble first" with a probability of 2/5:

P(A) = 2/5

And Event B is "get a Blue Marble second" ... but for that we have 2 choices:

If we got a Blue Marble first the chance is now 1/4

If we got a Red Marble first the chance is now 2/4

So we have to say which one we want, and use the symbol "|" to mean "given":

P(B|A) means "Event B given Event A"

In other words, event A has already happened, now what is the chance of event B?

P(B|A) is also called the "Conditional Probability" of B given A.

And in our case:

P(B|A) = 1/4

So the probability of getting 2 blue marbles is:

And we write it as

"Probability of event A and event B equals

the probability of event A times the probability of event B given event A"

Let's do the next example using only notation:

Example: Drawing 2 Kings from a Deck

Event A is drawing a King first, and Event B is drawing a King second.

For the first card the chance of drawing a King is 4 out of 52 (there are 4 Kings in a deck of 52 cards):

P(A) = 4/52

But after removing a King from the deck the probability of the 2nd card drawn is less likely to be a King (only 3 of the 51 cards left are Kings):

P(B|A) = 3/51

And so:

P(A and B) = P(A) x P(B|A) = (4/52) x (3/51) = 12/2652 = 1/221

So the chance of getting 2 Kings is 1 in 221, or about 0.5%

Finding Hidden Data

Using Algebra we can also "change the subject" of the formula, like this:

Start with: P(A and B) = P(A) x P(B|A)

Swap sides: P(A) x P(B|A) = P(A and B)

Divide by P(A): P(B|A) = P(A and B) / P(A)

And we have another useful formula:

"The probability of event B given event A equals

the probability of event A and event B divided by the probability of event A

Example: Ice Cream

70% of your friends like Chocolate, and 35% like Chocolate AND like Strawberry.

What percent of those who like Chocolate also like Strawberry?

P(Strawberry|Chocolate) = P(Chocolate and Strawberry) / P(Chocolate)

0.35 / 0.7 = 50%

50% of your friends who like Chocolate also like Strawberry

Big Example: Soccer Game

You are off to soccer, and want to be the Goalkeeper, but that depends who is the Coach today:

with Coach Sam the probability of being Goalkeeper is 0.5

with Coach Alex the probability of being Goalkeeper is 0.3

Sam is Coach more often ... about 6 out of every 10 games (a probability of 0.6).

So, what is the probability you will be a Goalkeeper today?

Let's build a tree diagram. First we show the two possible coaches: Sam or Alex:

The probability of getting Sam is 0.6, so the probability of Alex must be 0.4 (together the probability is 1)

Now, if you get Sam, there is 0.5 probability of being Goalie (and 0.5 of not being Goalie):

If you get Alex, there is 0.3 probability of being Goalie (and 0.7 not):

The tree diagram is complete, now let's calculate the overall probabilities. Remember that:

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

Here is how to do it for the "Sam, Yes" branch:

(When we take the 0.6 chance of Sam being coach and include the 0.5 chance that Sam will let you be Goalkeeper we end up with an 0.3 chance.)

But we are not done yet! We haven't included Alex as Coach:

An 0.4 chance of Alex as Coach, followed by an 0.3 chance gives 0.12

And the two "Yes" branches of the tree together make:

0.3 + 0.12 = 0.42 probability of being a Goalkeeper today

(That is a 42% chance)

Check

One final step: complete the calculations and make sure they add to 1:

0.3 + 0.3 + 0.12 + 0.28 = 1

Yes, they add to 1, so that looks right.

Friends and Random Numbers

Here is another quite different example of Conditional Probability.

4 friends (Alex, Blake, Chris and Dusty) each choose a random number between 1 and 5. What is the chance that any of them chose the same number?

Let's add our friends one at a time ...

First, what is the chance that Alex and Blake have the same number?

Blake compares his number to Alex's number. There is a 1 in 5 chance of a match.

As a tree diagram:

Note: "Yes" and "No" together makes 1

(1/5 + 4/5 = 5/5 = 1)

Now, let's include Chris ...

But there are now two cases to consider:

If Alex and Blake did match, then Chris has only one number to compare to.

But if Alex and Blake did not match then Chris has two numbers to compare to.

And we get this:

For the top line (Alex and Blake did match) we already have a match (a chance of 1/5).

But for the "Alex and Blake did not match" there is now a 2/5 chance of Chris matching (because Chris gets to match his number against both Alex and Blake).

And we can work out the combined chance by multiplying the chances it took to get there:

Following the "No, Yes" path ... there is a 4/5 chance of No, followed by a 2/5 chance of Yes:

(4/5) × (2/5) = 8/25

Following the "No, No" path ... there is a 4/5 chance of No, followed by a 3/5 chance of No:

(4/5) × (3/5) = 12/25

Also notice that when we add all chances together we still get 1 (a good check that we haven't made a mistake):

(5/25) + (8/25) + (12/25) = 25/25 = 1

Now what happens when we include Dusty?

It is the same idea, just more of it:

OK, that is all 4 friends, and the "Yes" chances together make 101/125:

Answer: 101/125

But here is something interesting ... if we follow the "No" path we can skip all the other calculations and make our life easier:

The chances of not matching are:

(4/5) × (3/5) × (2/5) = 24/125

So the chances of matching are:

1 - (24/125) = 101/125

(And we didn't really need a tree diagram for that!)

And that is a popular trick in probability:

It is often easier to work out the "No" case

(This idea is shown in more detail at Shared Birthdays.)

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