Tut 4

© Dr Raymond Teo, 2011 Probability Page 1 of 15

TUTORIAL 4: PROBABILITY

1. Say whether the following statement is true or false and briefly give your

reasons.

“If two events are independent, they must be mutually exclusive”

(Study guide – Sample Question)

Answer:

Two events are mutually exclusive if they cannot jointly occur, i.e. the probability of

their intersection is 0. In contrast, two events are independent if they are not related

to one another. The above statement is false as it is possible that two independent

events may not be mutually exclusive. To illustrate, Event A denotes good weather in

Singapore and Event B denotes a rise in the stock market. The two events are

independent because it is not expected that the weather has an impact on the stock

market. However, the two events need not be mutually exclusive as it is possible to

find situations that both conditions hold i.e. good weather and rise in stock market.

2. If X can take values of 1, 2 and 4 with P(X = 1) = 0.3, P(X = 2) = 0.5, P(X =

4) = 0.2, what are

a) ?)4P(X2≤

b) P(X>2|X is an even number)?

Answer to a)

The sample space given by the value of X = {1, 2, 4}.

The event that X2 is less or equal to 4 = {1, 2}

Since the events are mutually exclusive, 2) P(X 1)P(X )4P(X2=+==≤

= 0.3 + 0.5 = 0.8

Answer to b)

P(X>2|X is even) = even) is P(X

even) is X 2P(X ∩>

The event (X>2 ∩ X is even) = {4}. P(X>2 ∩ X is even) = 0.2

The event (X is even) = {2, 4}. P(X is even) = P(X = 2) + P(X = 4) = 0.5 +0.2 = 0.7

Therefore P(X>2|X is even) = 0.2/0.7 = 0.286


3. Write down and illustrate the use of probability of:

a) The addition rule

b) The multiplication rule.


Answer to a)

The addition rule of probability states that the probability of their union is the sum of

the individual probabilities minus the intersection of the probabilities, i.e.

B).P(AP(B)P(A)B)P(A ∩−+=∪

In a throw of dices, let A represent the event when an outcome is even {2,4,6} and B

represent the event when the outcome is 1, 2, or 3. P(A) = ½ and P(B) = ½. Since A

∩ B is denoted by the number 2, P(A ∩ B) = 1/6 = 0.1667. P(A ∪ B) = ½ + ½ -

0.1667 = 0.833.

Answer to b)

The multiplication rule of probability states that

)()|()( BPBAPBAP =∩

In a throw of dices, let A represent the event when an outcome is even {2,4,6} and B

represent the event when the outcome is 1, 2, or 3. P(B) = 1/2. As there is only 1

even number in B, the probability that A occurs given B has occurred is 1/3 = 0.333.

Hence P(A|B)P(B) = (0.333)(0.5) = 0.1667.

This can be verified by examining A ∩ B i.e. {2}. Since there is only one outcome

for

A ∩ B but 6 possible outcomes in a throw of a dice, P(A ∩ B) = 1/6 = 0.1667, which

is the same as before.

4. A fair die is thrown twice. What is the probability that:

i) A six turns up each time?

ii) Both numbers are odd?

iii) The sum of scores is 4?

(Year 2003 – Zone A)

Answer to i):

i) Total number of outcomes = 36

Events where six turns up each time: {(6,6)}

Number of events where six turns up each time = 1

P(six turns up each time) = 1/36

ii) Total number of outcomes = 36

Events where both number are odd: {(1,1,), (1,3), (1,5), (3,1), (3,3),

(3,5), (5,1), (5,3), (5,5,)}

Number of events where both are odd = 9

P(both numbers are odd) = 9/36 = 1/4


iii) Total number of outcomes = 36

Events where sum of two throws are 4 is represented by

{(1,3),(2,2),(3,1)}

Number of events where sum of two throws are 4 = 3

P(sum of two throws are 4) = 3/36 = 1/12

5. x can take values 1, 2, 3 and 5 with probability 0.1, 0.3, 0.4 and 0.2

respectively.

Find

i) E(x)

ii) The probability that x is odd given that x is less than 4.

iii) The probability that x2 > 8

Answer to i)

E(x) = 0.1*1 + 0.3*2 + 0.4*3 + 0.2*5

= 2.9

Answer to ii)

P(x=odd|x<4) = 4 x

4) x oddP(x

<

<∩=

Sample space representing {1,3} 4 x odd x =<∩= . Thus

0.5. 0.4 0.1 4)x oddP(x =+=<∩=

P(x<4) = 0.1 + 0.3 + 0.4 = 0.8.

Thus the required probability = 625.00.8

0.5=

Answer to iii)

The outcomes where this will happen is denoted by E = {3, 5}

Thus, the required probability is 0.4 + 0.2 = 0.6.


6. A fair dice is thrown twice. What is the probability that:

i) A one turns up each time?

ii) The sum of the scores is 10?

iii) Both numbers are even?

(Year 2003 – Zone B)

Answer to i)

Total number of events = 36

Event where one turns up each time: {(1,1)}

Number of events where one turns up each time = 1

P(one turns up each time) = 1/36

Answer to ii)

Total number of outcomes = 36

Events where sum is 10: {(4,6), (5,5), (6,4)}

Number of events where sum is 10 = 3

P(sum is 10) = 3/36 = 1/12

Answer to iii)


Events where both are even: {(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4),

(6,6)}

Number of events where both are even = 9

P(both are even) = 9/36 = 1/4

7. x can take values 1, 4, 5, 6 with a probability of 0.2, 0.4, 0.3 and 0.1

respectively. Find

i) E(x)

ii) The probability that x is odd, given that x is greater than 4

iii) The probability that x2 < 25

Answer to i)

E(x) = 0.2*1 + 0.4*4 + 0.3*5 +0.1*6

= 3.9.

Answer to ii)

P(x = odd|x >4) = 75.01.03.0

3.0

6) P(x 5) P(x

5) P(x

4)P(x

4) x and odd P(x =

+

=

=+=

==

>

>=

Answer to iii)

For this to happen, x = {1,4}.

P(x2 < 25) = P(x = 1) + P(x = 4) = 0.2 + 0.4 = 0.6


8. You toss a coin twice. Give

i) The probability that you get one head and one tail overall.

ii) The probability that you get a tail on the first throw and a head on the

second.

(Year 2004 – Zone A)

Answer to i)


Events where one is head and one is tail: {(H,T), (T,H)}

Total number of events where one is head and one is tail = 2

P(one is head and one is tail) = 2/4 = 1/2

Answer to ii)


Events where one is head and one is tail: {(H,T)}

Total number of events where head on the first throw and tail on second = 1

P(head on the first throw and tail on second) = 1/4

9. A student can enter a course either as a beginner (73%) or as a transferring

student (27%). It is found that 62% of beginners eventually graduate, and that

78% of transfers eventually graduate. Find:

a) the probability that a randomly chosen student is a beginner who

will eventually graduate,

b) the probability that a randomly chosen student will eventually

graduate,

c) the probability that a randomly chosen student is either a beginner

or will eventually graduate, or both.

Are the events èventually graduate’ and ‘enters as a transferring student’

statistically independent?

If a student eventually graduates, what is the probability that the student

entered as a transferring student?

If two entering students are chosen as random, what is the probability that not

only do they enter in the same way but that they also both graduate or fail?



Probability tree of student progress

Answer to a)

P(student is a beginner and will graduate) = P(beginner) P(graduate) = 0.4526

Answer to b)

P(student will eventually graduate) = P (beginner and graduates) + P (transfer

and graduates) = 0.4526 + 0.2106 = 0.6632

Answer to c)

P(student is either a beginner or will eventually graduate or both)

= 1 – P(student is a transfer and fails) = 1 – 0.0594 = 0.9406.

The two events èventually graduates’ and ènters as a transferring student’ is not

statistically independent as the probability of a student graduating depends on whether

or not the student is a transfer student.

P(transfer|graduate) = )P(graduate

graduate) P(transfer ∩= 3176.0

6632.0

2106.0=

P(both students are beginners and both graduate) = 0.4526 * 0.4526 = 0.2048

P(both students are beginners and both fail) = 0.2774 * 0.2774 = 0.0770

P(both students are transfers and both graduate) = 0.2106 * 0.2106 = 0.0444

P(both students are transfers and both fail) = 0.0594 * 0.0594 = 0.0035

Therefore probability that both students follow the same path = 0.2048 + 0.0770 +

0.0444 + 0.0035 = 0.3297.

P(beginner) = 0.73

P(transfer) = 0.27

P(graduate) = 0.62

P(fail) = 0.38

P(graduate) = 0.78

P(fail) = 0.22

P(beginner and

graduates) = 0.4526

P(beginner and

fails) = 0.2774

P(transfer and

graduates) = 0.2106

P(transfer and fails)

= 0.0594


10. A coffee machine may be defective because it dispenses the wrong amount of

coffee (C) and/or it dispenses the wrong amount of sugar (S).

The probabilities of these defects are:

P(C) = 0.05, P(S) = 0.04, P(C and S) = 0.01

What proportions of cups of coffee have:

a) at least one defect?

b) no defects?

(Study guide – Sample Questions)

Answer to a)

S)P(C - P(S) P(C) S) P(C rule, additive theApplying ∩+=∪

= 0.05 + 0.04 – 0.01 = 0.08

Answer to b)

P(no defects) = S) P(C - 1 S)P(C c∪=∪

= 1 – 0.08 = 0.92

11. You toss two fair dice.

i) What is the probability that both are sixes?

ii) You are now told that at least one of them shows a six, what is the

probability in this case that both are sixes?

iii) What is the probability that both numbers are odd?

(Zone B – 2005)

Answer to i)

Total number of events = 36

Event where one turns up each time: {(6,6)}

Number of events where one turns up each time = 1

P(one turns up each time) = 1/36

Venn Diagram of Defect in Machine

Coffee

dispensing

defect

Sugar

dispensing

defect


Answer to ii)

Events where at least one is a six: {(1,6), (2,6), (3,6), (4,6), (5,6), (6,6,), (6,1),

(6,2), (6,3), (6,4), (6,5)}

Number of events where at least one is six = 11

Number of events from this set where both are sixes = 1

P(both are six/at least one is six) = 1/11

Answer to iii)


Events where both number are odd: {(1,1,), (1,3), (1,5), (3,1), (3,3), (3,5),

(5,1), (5,3), (5,5,)}

Number of events where both are odd = 9

P(both numbers are odd) = 9/36 = 1/4

12. x takes the values 1, 3, 5 and 7 with the probabilities .1, .3, .5 and .1

respectively.

Find

i) The probability that x is an even number

ii) E(x)

iii) The probability that x2 > 20

(Zone B – 2005)

Answer to i)

Since the total probablities .1 + .3 + .5 + .1 = 1,

the probability is 0 that X is an even number

Answer to ii)

E(X) = 1 (.1) + 3 (.3) + 5 (.5) + 7 (.1)

= 4.2

Answer to iii)

P(x2 > 20) = P(x = 5) + P(x = 7)

= .5 + .1

= .6


13. A tennis player plays two sets against an opponent. He has a 50% chance of

winning the first set, and if he wins the first, he then has a 55% chance of

winning the second set. But it he loses the first, he has only a 40% chance of

winning the second set.

i) What is the probability of his winning both sets?

ii) If you are told he lost the second set, what is the probability he lost

both sets?

(Zone B – 2005)

Answer:

ii. P(lost both sets/lost second set)

= P(lost both sets ∩ lost second set)/P(lost second set)

= 0.300/(0.225 + 0.300)

= 0.571

14. An urn contains 3 balls coloured red, blue and green. These are drawn at

random one at a time, from the urn without replacement. If you are told that the green

ball was drawn before the red ball, what is the probability that

i) The red ball was drawn last?

ii) The blue ball was also drawn before the red ball?

(Zone B – 2005)

Probability Tree of Outcome:

P(winning) = 0.5

P(losing) = 0.5

First Set Second Set

P(winning) = 0.55

P(losing) = 0.45

P(winning) = 0.40

P(losing) = 0.60

P(winning first, winning second) = 0.275

P(winning first, losing second) = 0.225

P(losing first, winning second) = 0.200

P(losing first, losing second) = 0.300

P(winning first, winning second) = 0.5 x 0.55 = 0.275


Answer to i)

Answer to ii)

15. i) Two fair dices are thrown and you are told that the sum of their upturned

faces is equal to 7. What is the probability that neither face is equal to 6?

ii) Three balls are thrown at random into five bowls so that each ball has the

same chance of going into any bowl independently of wherever the other 2

balls fall. Determine the probability distribution of the number of empty

bowls.

(2006 – Zone B)

Answer

i)

first dice

Second dice

There are 6 possibilities for the dice to have a sum of 7.

1 2 3 4 5 6

1 1,1 2,1 3,1 4,1 5,1 6,1

2 1,2 2,2 3,2 4,2 5,2 6,2

3 1,3 2,3 3,3 4,3 5,3 6,3

4 1,4 2,4 3,4 4,4 5,4 6,4

5 1,5 2,5 3,5 4,5 5,5 6,5

6 1,6 2,6 3,6 4,6 5,6 6,6

R B G

R G B

B R G

B G R

G R B

G B R

All possible sets

Number of occurrencesof G before R , N = 3

Number of occurrencesof R last = 2

P( Red last / Green before Red) = 2/3

i)

R B G

R G B

B R G

B G R

G R B

G B R

All possible sets

Number of occurrencesof G before R , N = 3

Number of occurrencesof B before R = 2

P( Blue before Red / Green before Red) = 2/3

ii)


Out of these 6 possibilities, there are 4 possibilities that neither of the

faces are 6.

Hence, P(neither face is equal to 6/sum of upturned faces equals 7)

= 4/6 = 0.67

ii) P(0 bowls empty) = 0 [an impossible situation since there are 3 balls]

P(1 bowl is empty) = 0 [an impossible situation since there are 3 balls]

P(2 bowls empty) = P(first ball occupying any bowl and second ball

occupying a bowl that is empty and the third ball occupying a bowl that is

empty)

= 1 * 4/5 * 3/5 = 12/25

P(4 bowls empty) = P(first ball occupying any bowl and second ball

occupying the same bowl as the first and third ball occupying the same bowl

as the first)

= 1 x 1/5 * 1/5 = 1/25

P(5 bowls empty) = 0 [an impossible situation since the balls have to occupy

at least one bowl)

P(3 bowls empty) = 1 – P(1 bowl empty) – P(2 bowls empty) – P(4 bowls

empty) – P(5 bowls empty)

= 1 - 0 – 12/25 – 1/25 – 0 = 12/25

Alternative working for 3 empty bowls

1st ball – can go into any bowls (1)

2nd

ball –

Case 1: same bowl as 1st ball (1/5)

3rd

ball – any other bowl that is empty (4/5)

Case 2: not the same bowl as the first ball (4/5)

3rd

ball – same bowl as either 1st or 2

nd ball (2/5)

Total probability = (1/5 x 4/5) + (4/5 x 2/5) = 12/25

Probability distribution:

Empty bowls, x = 0 1 2 3 4 5

P(x) = 0 0 12/25 12/25 1/25 0


16. State whether the following are possible or not.

i) Only John or Peter may get the end of the year prize. The probability John

gets it is 0.4. That for Peter is 0.7

ii) The correlation coefficient for the relationships between the number of

years a woman spent at school and the number of children she has is minus

3.5.

iii) The variance of the results is minus 64

iv) The larger a sample, the smaller the variation in its sample mean.

(Zone B – 2006)

Answer:

i) This statement is not possible. As the two events are mutually exclusive

(only John or Peter may get the prize), the sum of the two probabilities

must equal one.

ii) Correlation coefficients can only range between -1.0 and 1.0. Hence the

statement is not possible.

iii) Variance can only take a positive value. Hence the statement is not

possible.

iv) The statement is possible. This follows from the central limit theorem,

which states the variance of the sampling distribution of means is equal to

the variance of the population from which the samples were drawn divided

by the size of the samples.

17.

(Zone A – 2006)

Answer

i) This statement is not possible. As the two events are mutually exclusive

(only Angela or Ester may get the prize), the sum of the two probabilities

must equal one.

ii) Correlation coefficients can only range between -1.0 and 1.0. Hence the

statement is not possible.


iii) This is possible as the variance can take on any positive value.

iv) The statement is not possible. This follows from the central limit theorem,

which states the variance of the sampling distribution of means is equal to

the variance of the population from which the samples were drawn divided

by the size of the samples. Therefore if the sample size increases, the

variable in its sample mean should be reduced.

16.

(Zone A – 2006)

Answer

i

2

1

40

20

40

8

40

12

40

16

BlondeBluePBlondePBluePBlondeBlueP

==−+=

∩−+=∪ )()()()(

ii.

2

1

2

11

)BlondeBlue(P1)blonenor eyedblue Neither(P

=−=

∪−=−

iii.

5

1

40

8

40

8

40

16

)BlondeBlue(P)Blue(P)blondenot but eyedblue(P

==−=

∩−=−


17.

(Zone A – 2007)

Answer:

The events are independent, so P(rain)P(umbrella) = P(rain and umbrella).

Since P(rain) = ¼ and P(rain and umbrella) = 5/16, this would mean

P(umbrella) exceeds 1, which is not possible.

18.

(Zone A – 2007)

Answer:

i) Total possible outcomes = [TTT,THT,TTH,HTT,THH,HTH,HHT,HHH]

Number of possible outcomes = 8

Outcomes with exactly two tails = [THT,TTH,HTT]; Number of outcomes = 3

P(exactly two tails) = 3/8

ii) Outcomes where at least one is a tail = [TTT,THT,TTH,HTT,THH,HTH,HHT]

Number of outcomes = 7

Number of outcomes where all are tails = 1 i.e. [TTT]

P(all are tails/at least one is a tail) = 1/7

19.

Two fair dice are thrown. What is the probability of at least one odd number?

What is the probability of this if four fair dice are thrown?

(Zone A – 2008)

Answer

i) Total possible number of outcomes with throw of two dices = 36

Total outcomes in event with at least one odd number = 27 {need to show}

P(at least one odd number) = 27/36 = 0.75


ii) Total number of outcomes = 6 x 6 x 6 x 6 = 1296

Number of outcomes with all even numbers: 3 x 3 x 3 x 3 = 81

Number of outcomes with at least one odd number = 1296 – 81 = 1215

P(at least one odd number ) = 1215/1296 = 0.9375

20.

In a manufacturing plant, machine A produces 10% of a certain product and

machine B produces 90% of this product. Of the production by machine A,

10% are defective; for machine B the defective rate is 5%. If a product is

selected at random from one of the machines, what is the probability that it is

defective?

If a company inspector, sampling production at random, discovers a defective

product, what is the chance that it came from Machine B?

(Zone A – 2008)

Answer:

{Draw a probability tree for this answer}

P(defect) = P(A and defective) = 0.1 x 0.1 = 0.01

P(defect) = P(B and defective) = 0.9 x 0.05 = 0.045

Since the two outcomes are mutually exclusive,

P(defective) = 0.01 + 0.045 = 0.055

P(B/defective) = P(B and defective)/P(defective)

= (0.9 x 0.05)/0.055

= 0.8182

grouper

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