The probability of Davis getting a merit and above for his Probability NECA exam Davis only pays...

The probability of Davis getting a merit and above for his Probability NECA exam

Davis only pays attention in class 2/5 of the time. If he pays attention the probability of him getting not achieved is 1/6, getting achieved is 4/6, getting merit and above is 1/6. If he doesn’t pay attention, the probability of him getting not achieved is 3/5, achieved 3/10, merit and above 1/10.

A basket ball team plays 2/5 of their game at home. If they are playing at home the probability of their winning is 0.7, but if they are playing away, the probability of a win is only 0.5. calculate the probability that they win their next game if it is not known whether the game is going to be played at home or away.

A blind person has tins of baked beans and of spaghetti. There are 6 tins of baked beans and 4 tins of spaghetti. The person opens three of the tins at random. Calculate the probability that two of the three tins are spaghetti and the other is baked beans.

R

G

B

R

G

B

R

G

B

Eg marbles in a bag 3 red, 2blue, 5 green. We take out one and then DON’T put it back and then take out another.

R

G

B

Out comesRR

RB

RGBR

BB

BGGR

GB

GG5 4 20

10 9 90

3 2 6

10 9 90

3 2 6

10 9 90

3 5 15

10 9 90

2 3 6

10 9 90

2 1 2

10 9 90

2 5 10

10 9 90

5 3 15

10 9 90

5 2 10

10 9 90

What is the probability at least one red marble is drawn?

*

*

*

*

*

A deck of 52 cards contains 13 of each suit: Hearts, diamonds, spades, and clubs calculate the probability that 4 successive draws without replacement produce

a) 4 Hearts

13

52

13

52

39

52

39

52

H

NH

12

51

12

51

39

51

39

51

H

NH

11

50

11

50

39

50

39

50

H

NH

10

49

10

49

39

49

39

49

H

NH

We do not need to complete the whole tree but only the branch we need

13 12 11 10(4 )

52 51 50 4917160

64974000.00264

P H

An Airline operates 3 aircraft. Each one could develop a technical problem and be grounded independently of the others. The probabilities for each on any given day are, 0.1,0.05,0.02.

Complete this table

x 0 1 2 3

P(X=x) 0.0001

An Airline operates 3 aircraft. Each one could develop a technical problem and be grounded independently of the others. The probabilities for each on any given day are, 0.1,0.15,0.02.

F

G

F

G

Aircraft 1

Aircraft 2

Aircraft 3 F

G

F

G

F

G

F

G

F

G

FFF

FFG

FGF

FGG

GFF

GFG

GGF

GGG

0.0001

0.00490.0019

0.09310.009

0.0441

0.0171

0.8379

x 0 1 2 3

P(X=x) 0.0001

F

G

F

G

Aircraft 1

Aircraft 2

Aircraft 3 F

G

F

G

F

G

F

G

F

G

FFF

FFG

FGF

FGG

GFF

GFG

GGF

GGG

0.0001

0.00490.0019

0.09310.009

0.0441

0.0171

0.8379

0.8379 0.1543 0.0077

Conditional probability

Conditional probability

If a probability is influenced by the occurrence of another event then the

probability is conditionalEx. What is the probability of students passing the second test if they passed the first test?

What is the probability of selecting a left-handed student given that the student has green eyes?

What is the probability that a student is absent given that today is Friday?

How to identify if it’s a Conditional probability?

Look for key words

“IF” and “GIVEN”

How to represent conditional probability?

means the probability of B given A has occuredBP A

means the probability of A given B has occuredAP B

The Condition is always at the bottom

What is the probability of selecting a left-handed student given that the student has green eyes?

P ( L / G )

How to calculate a conditional probability?

Generally given as

( )( / )

( )

P A BP A B

P B

The probability of A and Boccurring at the same time

35% of shoppers buy coffee, 25% buy tea, 5% buy both

Given that a shopper buys coffee what is the probability that he buys tea?

( )( / )

( )

P A BP A B

P B

P(C)=0.35P(T)=0.25P(Both)=0.05 P(T/C)= P(T∩C) P(C)= 0.05/0.35=0.143

A tour company find that tourists on its coaches are 60% Australian, 30% American, and 10% New Zealanders. The proportion of each nationality who are women is 4/5 , 3/5 ,

2/3.

a) Calculate the probability that a tourist selected at random is a women?

b) Given that a tourist is a woman, what is the probability she is Australian?

( ) 0.48 0.18 0.0667

0.7267(4 )

P W

sf

( )( / )

( )

P Au WP Au W

P W

0.48

0.72670.6606

Combining probability tree with conditional probability

A scientist plants 7 seeds in plot A, 5 seeds in plot B, and 8 seeds in plot C. The probability for germination of the seeds in each plot are 0.6, 0.8 and 0.4, respectively.

a) Draw a probability tree and use it to calculate the probability that a seed chosen at random germinates.

b) Given that a seed has germinated, calculate the probability that it came from plot B.

Sometimes “if” and “given” is not used in conditional probabilities.

What do you do?Remember:

Conditional probability use a reduced sample space.

The probability that a person selected at random in New Zealand has green eyes

As a oppose to

The probability that a person randomly selected from brown haired people in New Zealand has green eyes.

Venn Diagram

The contents of the whole diagram must add to 1

Venn Diagrams

A Venn diagram can show the relationship between 2 events A and B.

The sample space is the entire rectangle

A B

The probability A occurs

P(A)

The probability B occurs

P(B)

The intersection

means both occured

P(A B)

The union, P(A B)

means either

A or B occurs

or both together

Probabilities in Venn Diagrams

A B

( ) 0.4

( ) 0.3

( ) 0.1

P A

P B

P A B

P(A)=0.4

P(B)=0.3

P(A not B)=0.3 P(B not A)=0.2

0.4

P(A B)=0.1

P(A U B) =0.3+0.2+0.1=0.6Or 0.3+0.4-0.1=0.6

Probability Rules:

( ) ( ) ( ) ( )P A B P A P B P A B

The intersection got added twice so we have to remove it

( ) ( ) ( ) ( )P A B P A P B P A B

We can rearrange this

Examples Ex 1.2#3a)

C D

If P(C)=0.8, P(D)=0.6 and P(C D)=0.5,

Calculate P(C D)

0.8

0.6

0.5

0.1

Now we will use the formula

P(C D)=P(C)+P(D) P(C D)

=0.8+0.6 0.5

0.9

Eg from Theta Ex 11.5 #11The probability of purchases before Easter:

Shopper has: hot cross buns:0.39

Easter eggs :0.46

Neither:0.23

a) What is the probability that a shopper has both hot cross buns and Easter eggs?

Eg from Theta Ex 11.5 #11The probability of purchases before Easter:

Shopper has: hot cross buns:0.39

Easter eggs :0.46

Neither:0.23

b) What is the probability that a shopper has hot cross buns but not Easter eggs?

Eg from Theta Ex 11.5 #12The 6th form dean knows that of the 300 students in her

secondary school, the following numbers take commercial subjects

Business Studies 86Accounting 95Economics 109Business Studies and Accounting 23Accounting and Economics 38Economics and Business Studies 47All three 12

a) Draw a Venn diagram

b) If a student is chosen at random, what is the probability that they do not take any of these subjects?

Types of Events: Complementary Events

Complementary Events:If we consider an event to be AThen we consider any other event not A, we can write this A’

A

A’

Remember the total of the diagram must add to 1

( ) ( ') 1P A P A

( ') 1 ( )P A P A

Independent events

Types of Events: Mutually Exclusive Events

Mutually exclusive events

cannot both occur

they have nothing in common

the circles in the Venn diagrams do not over lap.

there is no intersection

If we consider the event A and event B

The probabilities can be added

B( ) ( ) ( )P A B P A P B

There is no intersection so we don’t have to subtract it

( ) 0P A B

A

Types of Events: Independent Events

If events are independent the occurrence of A has no effect on the occurrence of B

( ) ( ) ( )P A B P A P B

This is an important statement

If it is true, then events are independent

If events are independent, then it is true

We can use this to prove events are independent, or if they are independent then we can use it to find the probability that both occur

Examples from Ex 1.3#2The probability that an person is overweight is 1/3, the probability that a person is diabetic is 3/10. The probability that a person is overweight and has diabetes is 1/5. Are these events independent?If events are independent then

( ) ( ) ( )P A B P A P B

( ) ( ) ( )P O D P O P D

1 3 1

3 10 10

1 1events are not independent

10 5as

Examples from Ex 1.3 #9A business has two phone lines each with a probability of 0.01

of developing a fault. The probability that both develop a fault is 0.001

a) Are the events independent?

( ) ( ) ( )P A B P A P B

0.001 0.0001 events are not independentas

0.01 0.01 0.0001

Examples from Ex 1.3 #9A business has two phone lines each with a probability of 0.01

of developing a fault. The probability that both develop a fault is 0.001

b) Calculate the probability that both lines are available.

( ) 0.001P A B

0.01 0.01 0.001

0.019

( ) ( ) ( ) ( )P A B P A P B P A B

If both lines are available then there is not fault , so let’s consider the probability of any fault and subtract it from 1

1 0.019 0.981

The probability that both line are available is 0.981

Expected Value

• The probability distribution for the random variable Y is shown below.

a. Find the value of k b. Find E(Y) c. Find Var (Y) and SD(Y)

• Find the expected number of tails when two fair coins are tossed.

• A raffle has one prize of $50 and one of $20. One hundred tickets are sold. What would be the expected value (average return) on $1 ticket?

• In a game of "unders and overs", two dice are tossed. The gambler pays $2 to play and wins $10 if the total is 7. What is the expected gain to the gambler.

y 2 4 6 8P(Y=y) k 4k 2k 3k

X is a random variable with E(x)=5 and Var(x)=4

i) Calculate the value of the standard deviation of X. SD(x)

ii) Calculate E(5x+1)

iii) Calculate Var(2x+1)

iv) Calculate SD(2x+1)

Revision

Type 1:Probability without replacement

2 cards are drawn from a deck without replacement. What is the probability that both cards are greater than 3 and less than 8?

Type 2: Venn Diagram

The probability of owning a boat is 0.2. The probability of owning a car is 0.7. The probability of owning a boat or a car or both is 0.8.

Find the probability of a) owning both a car and a boatb) Neither a car or a boat.

Type 3: Conditional probability 50% of people who drive into a car park

drive a Japanese car. 10% drive a German car. 20% drive an American car. 3/5 of Japanese car drivers are female. 2/5 of the German car drivers are female. 1/4 of American car drivers are female. When a female drivers drives into the car park. Find the probability that she’s in a Japanese car.

Type 4: Probability without replacement There are a total of 8 marbles. 2 red marbles, 5 blue marbles and 1 green marbles. 3 marbles are picked out randomly.

What is the probability of a) getting 0 blue marbles? b) 1 blue marbles? c) 2 blue marbles? d) 3 blue marbles?

Exercise

Type 4: Expected ValueA café has a total of 8 chicken sandwiches, 7

beef sandwiches and 5 tuna sandwiches. If one day it sells 4 sandwiches.

a) Complete the probability distribution table for selling 4 chicken sandwiches.

b) The expected number of chicken sandwiches sold

c) Then standard deviation of the number of chickens sandwiches sold

X 0 1 2 3 4 5P(X=x)

There are 120 raffles tickets which has 4 prizes of $150. How much should each ticket cost if a profit of $ 400 is to be made?

Type 5:Var(ax+by)=a2Var(x)+b2Var(y)

The weights of apples sold per day have mean 112.4kg and standard deviation 8.6kg. The weights of pears sold per day have mean 75.6 kg and standard deviation 9.2kg. If apples sell for $2 per kg and pears $3 per kg. find the probability the daily takings for apple and pear sales are between $400 and $500.

Exercise

Type 6:Combination and permutation

A café sells cakes and sandwichesCake: chocolate, toffee, blueberry, vanilla, fruitSandwiches: chicken, beef, salmon, tuna

A customer is going to buy 3 items at randoma) How many selections are possible?b) How many if the customer is not allowed to

have both cakesc) How many if exactly 2 must cakes.

Venn Diagram (Revision)Venn Diagram (Revision) The probability of being married is

0.65, the probability of owning a home is 0.56. The probability of being married and owning a home is 0.51. Find the probability of owning your own home or being married.

Probability Revision (Venn Probability Revision (Venn diagram)diagram)

50 students. 20 study calc, 26 study stats.

• If 40 students study either calc or stats. What’s the probability that a randomly chosen student does both?

• If 10 students study both. What’s the probability that a randomly chosen student does neither of the two?

Venn diagramVenn diagram

120 people in total. 50 people swim, 60 people play tennis. 100 play either swim or play tennis. what is the probability that a person plays tennis and swims?

Expected Value If the total on the dice is 11 or 12,

the person wins $3. If the total on the dice is 7 the person wins $1 and if the total on the dice is $4 or less the person wins $2.

a) Draw a probability distribution table and calculate the expected amount.

b) Calculate the variance of the distribution

End of the year revision

Probability

• Venn diagram

P(AUB)=P(A)+P(B)-P(A∩B)

Rearrange it

P(A∩B)=P(A)+P(B)-P(AUB)

Types of Events: Complementary Events

A’

( ) ( ') 1P A P A

( ') 1 ( )P A P A

A

( ) ( ) ( )P A B P A P B A B

Types of Events: Mutually Exclusive Events

Types of Events: Independent Events

If events are independent the occurrence of A has no effect on the occurrence of B

( ) ( ) ( )P A B P A P B

2004 Achieved

2004 Achieved

2004 Merit

2005 Achieved

2005 Achieved

Conditional probability

If a probability is influenced by the occurrence of another event then the

probability is conditionalEx. What is the probability of students passing the second test if they passed the first test?

What is the probability of selecting a left-handed student given that the student has green eyes?

What is the probability that a student is absent given that today is Friday?

How to calculate a conditional probability?

Generally given as

( )( / )

( )

P A BP A B

P B

The probability of A and Boccurring at the same time

Condition is always atThe bottom

2005 Merit

2005 Achieved

Expected ValueX 2 4 6 8

P(X=x) 0.2 0.3 0.4 0.1

Expected Value=Variance=Standard Deviation=

X is a random variable with E(x)=5 and Var(x)=4

i) Calculate the value of the standard deviation of X. SD(x)

ii) Calculate E(5x+1)

iii) Calculate Var(2x+1)

iv) Calculate SD(2x+1)

2005 Achieved

2006 Merit

2004 Merit

2005 Merit

2005 Merit

2006 Excellence

2005 Merit

2006 Merit