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

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.
5
0.
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.
Out comes 3 2 6
 
*
RR
R
)
10 9 90
What is the probability
9
(2/
3 2 6
at least one red marble
 
RB
*
)
9
/
2
(
is drawn?
10 9 90
B
(5/9
)
)
R
0
3 5 15
1
/
 
3
(
RG
*
3
10 9 90
G
0.
R BR * 2  3  6
9)
(3/
10 9 90
0.2 (2/10)
(1/9)
2 1 2
B
 
(5/
BB
B
9)
10 9 90
2 5 10
G BG 10  9  90
)
R GR * 5  3  15
(3/9
10 9 90
(2/9)
5 2 10
G
(4/
B
 
GB
9)
10 9 90
5 4 20
G
GG 10  9  90
0)
/1
(5
10
49
39
49
A deck of 52 cards contains 13 of each suit: Hearts, diamonds, spades,
11
and clubs
calculate the probability that 4 successive draws without
50
replacement
produce
39
50
12
51
39
51
a) 4 Hearts
10
49
11
50
H
39
49
12
51
13
52
H
39
50
39
51
NH
NH
13 12 11 10
P(4 H )    
52 51 50 49
17160

6497400
NH
 0.00264
NH
H
39
52
H
We do not need to
complete the whole tree
but only the branch we
need
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
P(X=x)
0
1
2
3
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.
Aircraft 2
Aircraft 1
Aircraft 3
0.02
5
0.0
0.1
F
F
0.98
0.02
0.9
5
G
0.98
0.02
0.9
G
5
0.0
F
0.9
5
0.98
0.02
G
0.98
F
FFF
G
FFG 0.0049
FGF 0.0019
F
G
F
0.0001
FGG 0.0931
GFF 0.009
G
GFG 0.0441
F GGF 0.0171
G GGG
0.8379
x
P(X=x)
0
1
2
0.8379
0.1543
0.0077
Aircraft 2
Aircraft 1
Aircraft 3
0.02
5
0.0
0.1
F
F
0.98
0.02
0.9
5
G
0.98
0.02
0.9
G
5
0.0
3
0.0001
F
0.9
5
0.98
0.02
G
0.98
F
FFF
G
FFG 0.0049
FGF 0.0019
F
G
F
0.0001
FGG 0.0931
GFF 0.009
G
GFG 0.0441
F GGF 0.0171
G GGG
0.8379
Conditional probability
Conditional probability
If a probability is influenced by the
occurrence of another event then the
probability is conditional
Ex.
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?
 A means the probability of B given A has occured
P  A  means the probability of A given B has occured
B
P 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  B)
P( A / B) 
P( B)
The probability of A and B
occurring at the same time
P( A  B)
P( A / B) 
P( B)
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(C)=0.35
P(T)=0.25
P(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?
P (W )  0.48  0.18  0.0667
 0.7267(4 sf )
P( Au  W )
P( Au / W ) 
P(W )
0.48

0.7267
 0.6606
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
Venn Diagrams
A Venn diagram can show the relationship between 2 events A and B.
The
probability
A occurs
P(A)
A
B
The
sample
space
The
union,
P(Aisthe
B)entire rectangle
The
probability
B occurs
P(B)
The intersection
means of
either
The contents
the whole diagram must
add to
1 occured
means
both
A or B occurs
P(A  B)
or both together
Probabilities in Venn Diagrams
P( A)  0.4
P( B)  0.3
P(A  B)=0.1
P( A  B)  0.1
0.4
B
A
P(B)=0.3
P(A)=0.4
P(A not B)=0.3
P(B not A)=0.2
P(A U B) =0.3+0.2+0.1=0.6
Or 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


We can rearrange this
P( A  B)  P( A)  P( B)  P( A  B)

Examples Ex 1.2
#3a)
If P(C)=0.8, P(D)=0.6 and P(C  D)=0.5,
Calculate P(C  D)
0.1
D
C
0.8
0.5
0.6
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 #11
The 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 #11
The 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 #12
The 6th form dean knows that of the 300 students in her secondary
school, the following numbers take commercial subjects
Business Studies
86
Accounting
95
Economics
109
Business Studies and Accounting
23
Accounting and Economics
38
Economics and Business Studies
47
All 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 A
Then we consider any other event not A, we can write this A’
Remember the total
of the diagram must
add to 1
P( A)  P( A ')  1
P( A ')  1  P( A)
A’
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
P( A  B)  0
If we consider the event A and event B
The probabilities can be added
P( A  B)  P( A)  P( B)
There is no intersection so
we don’t have to subtract it
A
B
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#2
The 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?
P( A  B)  P( A)  P( B)
If events are independent then
P(O  D)  P(O)  P( D)
1 3
1
 
3 10 10
as
1 1
  events are not independent
10 5
Examples from Ex 1.3 #9
A 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)
0.01 0.01  0.0001
as 0.001  0.0001 events are not independent
Examples from Ex 1.3 #9
A 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.
If both lines are available then there is not fault , so let’s consider the
probability of any fault and subtract it from 1
P( A  B)  P( A)  P( B)  P( A  B)
P( A  B)  0.001
 0.01  0.01  0.001
 0.019
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.
y
P(Y=y)
2
k
a. Find the value of k
b. Find E(Y)
c. Find Var (Y) and SD(Y)
4 6
4k 2k
8
3k
• 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.
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 boat
b) 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 Value
A 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.
X
0
1
2
3
4
5
P(X=x)
b) The expected number of chicken sandwiches
sold
c) Then standard deviation of the number of
chickens sandwiches sold
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 sandwiches
Cake: chocolate, toffee, blueberry, vanilla, fruit
Sandwiches: chicken, beef, salmon, tuna
A customer is going to buy 3 items at random
a) How many selections are possible?
b) How many if the customer is not allowed to
have both cakes
c) How many if exactly 2 must cakes.
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 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 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’
A
P( A)  P( A ')  1
P( A ')  1  P( A)
Types of Events:
Mutually Exclusive Events
P( A  B)  P( A)  P( B)
A
B
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 conditional
Ex.
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  B)
P( A / B) 
P( B)
The probability of A and B
occurring at the same time
Condition is always at
The bottom
2005 Merit
2005 Achieved
Expected Value
X
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