Data Interpretation & Probability

Conditional Probability

A dice is rolled. Find the probability an even number will show on a single roll.
Recall    P(A die is even) =

What if the die were weighted toward the odd numbers? Then .5 is not the probability.

What is the probability of snow in the mountains?
The probability of snow in the month of December, is different than the probability of snow in the month of July.

Counting Rule for Conditional Probability

If A and B are events in a finite sample space consisting of equally likely outcomes then,

Think about it:

A classroom has 40 students; 30 students are girls and 10 students are boys. One of the students received an 86% on the last test. If a student is selected at random, what is the probability the student selected recived the 86%?

The answer is 1/40.

Next, I tell you the student that received the 86% is a girl. Now if a student is selected at random, what is the probability the student selected received the score of 86%?

The answer is 1/30.

The sample space of 40 students is reduced to 30 students.    The 10 students that are boys, are no longer used to calculate the answer.    This is called conditional probability.    The original sample space is reduced to the conditional sample space.
The denominator of the fraction is now 30 instead of 40.

The notation P(A/B) translates to, what is the probability that A occurs, if you are given the fact that B has occurred.    B is the reduced sample space.

When translating a word problem:

P(A\B) = P(statement: what is the probability, statement: if or given)

Example 1:

Two fair dice are rolled.    What is the probability the sum is 7, if we are given that each die comes up less than 5?

Solution:
The event A = the sum is 7.
The event B = each die comes up less than 5.

Given the information, each die comes up less than 5, the reduced sample space can be calculated.

Always calculate the reduced sample space first.

The sample space for event B:
= {(1,1),(1,2),(1,3),(1,4)
   (2,1),(2,2),(2,3),(2,4)
   (3,1),(3,2),(3,3),(3,4)
   (4,1),(4,2),(4,3),(4,4)}

The second step is to determine the number of elements that is in event A and the reduced sample space B.
The intersection of A and B is as follows:

P(sum is 7\each die comes up less then 5) = P(A\B)

=

Therefore P(sum is 7\each die comes up less then 5) = 1/8.

Theorem:

(The Intersection Rule)

For any events A and B,

_

Example 2:

A poll of 100 voters found that 70% favored candidate A, and 30% favored candidate B.    However, the poll also found that 62% of those that favored A are unlikely to vote, while 90% of B's supporters will vote.    Who should win the election?
Solution:
Draw a tree diagram.

To find the probability that A will win, we are only concerned with those who voted.    Thus find    P(A\V)

P(V\A) means voters, given the fact that they came from the A branch as opposed to the B branch.

P(V) is all ways to have a voter.    Voters are produced from either the A branch or the B branch.

P(A\B) = .4963 implies A should lose by a small amount.

Therefore B should win by a very small margin.

Student #1:
A large corporation employs 1000 men and 1500 women.    A survey indicates 725 of the men are college graduates as are 1200 women.    A person is selected at random.    Find the probability that a person is a female given that the person is a college graduate.

.62 .81 .76 .45

Answer

Student #2:
A large corporation employs 1000 men and 1500 women.    A survey indicates 725 of the men are college graduates as are 1200 women.    A person is selected at random.    Find the probability that the person is a college graduate given that the person is a male.

.62 .71 .65 .73

Answer

Example 3:

A card is drawn at random from a standard deck of playing cards.

A)    What is the probability that the card is higher than a seven, given no information at all?

Let    A = The event the card is higher than a seven.
There are 28 cards out of 52 that are higher than a seven.

We want:

Therefore P(higher than a seven) = 7/13.

B)    What is the probability that the card is higher than a seven, given the card is not an ace?
Let    A= The card is higher than a seven.
Let    B = The card is not an ace.
We want:
P(A\B) = P(the card is higher then a seven\the card is not an ace)

The sample space for B is all cards from a standard deck except the Aces.
There are 48 cards in this set where,
C = clubs    S = spade    H = hearts    D = diamonds

B={2C,2S,2H,2D,...,KC,KS,KH,KD}

The denominator of the fraction is 48.

To calculate the numerator, find the number of elements that are in both A and B.

A = The card is higher than a 7.
B = {2C,2S,2H,2D,...,KC,KS,KH,KD}

{8C,8S,8H,8D,...,KC,KS,KH,KD}    28 - 4 = 24 cards

Since the reduced sample space contains 48 elements, 48 is the denominator.
Since the intersection of event A and the reduced sample space B contains 24 elements, 24 is the numerator.

P(A\B) =

Therefore P(higher than a seven\the card is not an Ace) = 1/2.

C)    What is the probability that the card is higher than a seven, given the card is a club?
Let    A = The card is higher than a seven.
Let    B = The card is a club.

We want:
P(A\B) = P(The card is higher then a seven\the card is a club)

The sample space for B is    B = {13 clubs}.
Since the reduced sample space contains 13 elements, the denominator of the fraction is 13.

To calculate the numerator, find the number of elements that are in both A and B.

{8C, 9C, 10C, JC, QC, KC, AC}    7 members

P(A\B) =

Therefore P(higher than a seven\the card is a club) = 7/13.

D)    What is the probability that the card is a club, given the card is higher than a 10?
Let    A = The card is a club.
Let    B= The card is higher than a 10.

We want:
P(A\B) = P(The card is a club\the card is higher than a 10)

The sample space for B is all cards higher than a 10    {J, Q, K, A}.
The set B contains 16 cards.

   {J Clubs, Q Clubs, K Clubs, A Clubs}
The set of A and B contains 4 cards.

P(A\B) =

Therefore P(The card is a club\the card is higher than a 10) = .25.

Student #3:
Two fair dice are rolled.    What is the probability that the sum rolled is at least eight, given that at least one die comes up 5?

.32 .11 .64 .75

Answer

Student #4:
Two fair dice are rolled. What is the probability that the sum rolled is at least eight, given that the same number appears on both dice?

.6 .5 .98 .25

Answer

Student #5:
Suppose that one jar contains 12 red and 8 blue marbles, and a second jar contains 7 red and 10 blue marbles.    If 1 marble from the first jar is transferred to the second jar, mix up the marbles, and then remove a marble from the second jar, what is the probability that this marble is red?

Note:
Jar 1: 12 R, 8 B    total=20
Jar 2: 7 R, 10 B    total=17

.87 .65 .42 .36

Answer

Example 4:

A new product is surveyed for possible distribution of the product.

Used Daily

Used Occasionally

Total

Male

142

258

400

Female

619

381

1,000

Total

761

639

1,400

Let    D = Used Daily
Let    O = Used Occasionally

Let    F = Female
Let    M = Male

Assume someone is selected at random, find the following:

A)    Find    P(M)

Solution:
P(M) is the probability that a male is selected.

B)    Find    P(F\D)

Solution:
P(F\D) is the probability that a female is selected given the fact that the person is a daily user.    The reduced sample space is all daily users, so divide by all way to be a daily user.

C)    Find    P(M\D)

Solution:
P(M\D) is the probability that a male is selected, given the fact that the person is a daily user.    The reduced sample space is daily user, so divide out by all way to be a daily user.

D)    Find    P(D\M)

Solution:
P(D\M) is the probability that a daily user is selected given the fact that the person is a male.    The reduced sample space is male, so divide out by all way to be a male.

E)    Find    P(F\O)

Students do this problem.    The answer is .6

Student #6:
A manufacturer of calculators found that 10% of the calculators produced had cosmetic defects and 2% had both cosmetic and malfunction defects.    What is the probability that one calculator selected at random malfunctions if it is known that it has a cosmetic defect?

.2 .6 .44 .88

Answer

Student #7:
At a certain college a survey of students taking Finite Mathematics and Statistics found that 25% dropped Finite, 30% dropped Statistics, and 10% dropped both courses.    If a person dropped Finite Math, what is the probability that the person also dropped statistics?

.3 .73 .6 .4

Answer

Student #8:
A family has three children.    Find the probability of two boys given that at least one is a boy.
.37 .53 .43 .24

Answer

Homework Problems

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Student #1:    The answer is A:

Solution:
There are 2500 in the sample space and 1925 graduates in the reduced sample space.

Let    G=graduate
F = female
M = male

The intersection of Females and Graduates is 1200.

Therefore the probability is .6234

Go Back to Lesson

Student #2:    The answer is D:

Solution:
The reduced sample space is males.    There are 1000 men.

The intersection of graduates and males is 725.

Therefore the probability is .725

Go Back to Lesson

Student #3:    The answer is C:

Solution:
Let A = sum is at least 8
Let B = at least one die is a 5

We want:
P(A\B) = P(sum is at least 8\at least one die is a 5)

B={(1,5)(2,5)(3,5)(4,5)(5,5)(6,5)(5,1)(5,2)(5,3)(5,4)(5,6)}

   {(3,5)(4,5)(5,5)(6,5)(5,3)(5,4)(5,6)}

P(A\B)=

Go Back to Lesson

Student #4:    The answer is B:
Solution:
Let A = sum is at least 8
Let B = same number appears on both dice

We want:
P(A\B) = P(sum is at least 8\same number appears on both dice)

B={(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)}

   {(4,4)(5,5)(6,6)}

P(A\B)= 3/6

.5

Go Back to Lesson

Student #5:    The answer is C:

Solution:
The first jar is straight forward.    Red 12/20 and blue 8/20.    There are two different possibilities when moving the marble to the second jar.    The marble could be red or the marble could be blue.

If the marble is red that is moved, then the second jar will have a probability of 8/18 of getting a red marble and 10/18 chance of drawing a blue.

If the marble is blue that is moved, then the second jar will have a probability of 7/18 of getting a red marble and 11/18 chance of drawing a blue.

The probability of drawing a red marble after the transfer is:

= .42

Go Back to Lesson

Student #6:    The answer is A:

Solution:
Let M = malfunction
C = cosmetic defect

Go Back to Lesson

Student #7:    The answer is D:

Solution:

Let    S = Students who drop Statistics

Let    F= Students who drop Finite

Go Back to Lesson

Student #8:    The answer is C:

Solution:

S={GGG,GGB,GBG,GBB,BBB,BBG,BGB,BGG}

B={GGB,GBG,GBB,BBB,BBG,BGB,BGG}

P(Two boys\At least one boy) =

Go Back to Lesson