- Probability

Probability deals with the likelihood of an event occurring. Given an event \(A\) the probability of \(A\) occurring is often denoted \(P(A)\) and is calculated as follows:

\(\displaystyle P(A)={\tt{number \:of \:times \:A \:occurs} \over \tt{number \:of \:possible \:outcomes}}\)

LESSON 1
What is the probability of drawing an Ace from a pack of 52 cards.
SOLUTION
There are 4 Aces in a deck of 52 cards, therefore
\(\displaystyle P(\tt{Ace})={4\over52}={1\over13}\)

SAMPLE SPACE

A sample space lists all of the possible outcomes of an experiment.
LESSON 1
A fair die is rolled.
(i) Construct the sample space.
(ii) Given that \(X\) represents the possible outcomes, state the corresponding probability values.
SOLUTION

LaTeX: X

1

2

3

4

5

6

LaTeX: P(X)

LaTeX: \displaystyle \frac{1}{6}

LaTeX: \displaystyle \frac{1}{6}

LaTeX: \displaystyle \frac{1}{6}

LaTeX: \displaystyle \frac{1}{6}

LaTeX: \displaystyle \frac{1}{6}

LaTeX: \displaystyle \frac{1}{6}

The first row represents the sample space and the second row represents the corresponding probabilities.

LESSON 2
State the sample space if 2 dice are rolled simultaneously.
Determine the probability of rolling a double.
SOLUTION

1

2

3

4

5

6

1

1, 1

1, 2

1, 3

1, 4

1, 5

1, 6

2

2, 1

2, 2

2, 3

2, 4

2, 5

2, 6

3

3, 1

3, 2

3, 3

3, 4

3, 5

3, 6

4

4, 1

4, 2

4, 3

4, 4

4, 5

4, 6

5

5, 1

5, 2

5, 3

5, 4

5, 5

5, 6

6

6, 1

6, 2

6, 3

6, 4

6, 5

6, 6

There are 6 doubles (highlighted in the table) and 36 possible combinations.
\(\displaystyle P(Double)={6\over36}={1\over6}\)

BASIC LAWS OF PROBABILITY

  • \(0≤P(A)≤1\)
  • \(P(A)=0\) implies that event A will definitely NOT occur.’
  • \(P(A)=1\) means that event A will definitely occur
  • \(P(A)=1-P(A')\) where \(P(A')\) represents the probability of A not occurring
  • The sum of all possible outcomes in a sample space is 1.

TREE DIAGRAMS

Tree diagrams show all the possible outcomes of an event and it allows the calculations of their probabilities.

LESSON 1

An experiment consists of selecting a ball from a bag and spinning a coin. The bag contains 5 red balls and 7 blue balls. A ball is selected at random from the bag, its colour is noted and then the ball is returned to the bag.

When a red ball is selected, a biased coin with probability \(\displaystyle {2\over3}\) of landing heads is spun.
When a blue ball is selected a fair coin is spun.
Complete the tree diagram below to show the possible outcomes and associated probabilities.

SOLUTION

NB: Each set of branches sum to 1

If \(\displaystyle P(Red)={2\over3}\) then \(\displaystyle P(R')=P(Blue)=1-{2\over3}={1\over3}\).

If Red ball is selected
\(\displaystyle ⟹ P(Heads)={2\over3}\)
\(\displaystyle ⟹P(Heads')=P(Tails)=1-{2\over3}={1\over3}\)
If Blue Ball is selected
\(\displaystyle ⟹P(Heads)={1\over2}\)
\(\displaystyle ⟹P(Tails)={1\over2}\)

LESSON 2
A bag contains 9 blue balls and 3 red balls. A ball is selected at random from the bag and its colour is recorded. The ball is not replaced. A second ball is selected at random and its colour is recorded.

Draw a tree diagram to illustrate all the possible outcomes and associated probabilities.

SOLUTION

One the first draw \(\displaystyle P(Blue)={9\over12}\) and \(\displaystyle P(Red)={3\over12}\)
One the second draw we have two scenarios:
SCENARIO 1 – First ball was Blue
We would have removed one blue ball so there are now 8 blue balls and 11 balls in total
\(\displaystyle P(Blue)={8\over11}\) and \(\displaystyle P(Red)={3\over11}\)
SCENARIO 2 – First ball was Red
We would have removed one red ball so there are now 2 red balls and 9 balls in total
\(\displaystyle P(Red)={2\over11}\) and \(\displaystyle P(Blue)={9\over11}\)


MUTUALLY EXCLUSIVE EVENTS

A marble is randomly selected from a bag containing 2 red marbles and 3 blue marbles. Can this marble be red and black at the same time? Of course not! These events are mutually exclusive.
Two events \(A\) and \(B\) are mutually exclusive if they both cannot occur at the same time i.e. \(A∩B=0\).

\(P(A \:or\: B)=P(A∪B)=P(A)+P(B)\)
LESSON 1
A pack of Skittles contains 10 green, 16 yellow, 17 orange, 18 red and 13 purple. Calculate the probability of drawing
(a) Green or red skittle
(b) Red or purple skittle
SOLUTION
(a) There are 74 skittles in the pack. There are 28 skittles that are either green or red
\(\displaystyle P(G∪R)={28\over74}={14\over37}\)
Using the addition rule:
\(P(G∪R)=P(G)+P(R)\)
\(\displaystyle ={10\over74}+{18\over74}\)
\(\displaystyle ={28\over74}\)
(b) There are 31 skittles that are either red or purple
\(\displaystyle P(R∪Pur)={31\over74}\)
Using the addition rule:
\(P(R∪P)=P(R)+P(pur)\)
\(\displaystyle ={18\over74}+{13\over74}\)
\(\displaystyle ={31\over74}\)

LESSON 2
The events \(A\) and \(B\) are such that
\(P(A)=0.6,\:\:\:\:\: P(B)=0.2\)
Evaluate \(P(A∪B)\) when \(A, B\) are mutually exclusive.
SOLUTION
Since \(A\) and \(B\) are mutually exclusive
\(P(A∪B)=P(A)+P(B)\)
\(P(A∪B)=0.6+0.2\)
\(P(A∪B)=0.8\)

NON - MUTUALLY EXCLUSIVE EVENTS

If a card is randomly drawn from a deck of 52 cards, is it possible to get a heart and a King? The answer is yes; you would just have to draw the King of Hearts. Since these two events (drawing a Heart and drawing a King) have a point of intersection, so to speak, they are said to be non – mutually exclusive events.
Two events \(A\) and \(B\) are said to be mutually non – exclusive events if both the events \(A\) and \(B\) have at least one common outcome between them.

\(P(A∪B)=P(A)+P(B)-P(A∩B)\)

LESSON 1
In a class of 32 students all of the students play either football or tennis. 15 play football and 21 play tennis. What is the probability that a student chosen at random plays both football and tennis?
SOLUTION
\(P(F∪T)=P(F)+P(T)-P(F∩T)\)
\(\displaystyle {32\over32}={15\over32}+{21\over32}-P(F∩T)\)
\(\displaystyle P(F∩T)={15\over32}+{21\over32}-{32\over32}\)
\(\displaystyle P(F∩T)={5\over32}\)

LESSON 2
If a fair die is rolled what is the probability of obtaining a number less than 4 or an odd number?
SOLUTION We first of all note that the numbers less than 4 are {1, 2, 3} and the odd numbers are {1, 3, 5}. This will be now illustrated on a Venn diagram.
\(P(ODD∪<4)=P(ODD)+P(<4)-P(ODD∩<4)\)
\(\displaystyle ={1\over2}+{1\over2}-{2\over6}\)
\(\displaystyle ={2\over3}\)

CONDITIONAL PROBABILITY

\(P(A│B)\) is the conditional probability that event \(A\) occurs given that event \(B\) has already occurred.
\(\displaystyle P(A│B)={P(A∩B)\over P(B)}\)

Let’s consider the Venn diagram where \(H\) represents the students who study History and \(G\) represents the students who study Geography.

\(\displaystyle P(H│G)={6\over 17}\). This states that there are 6 people out of the 17 Geography students who study History.
\(\displaystyle P(G│H)={6\over21}\). This states that there are 6 people out of the 21 History students who study Geography.

LESSON 1
A bag contains 3 red marbles and 8 blue marbles. Marbles are drawn at random without replacement. What is the probability that the second marble drawn is blue given that the first marble drawn is red.
SOLUTION
If a red marble is drawn first without replacement there are now 10 marbles to choose from – 2 red and 8 blue.
\(\displaystyle P(B│R)={P(2nd \:blue \:∩\:1st \:red)\over P(1st \:red)}\)
\(\displaystyle ={{8\over10}×{3\over11}\over {3\over11}}\)
\(\displaystyle ={8\over10}\)
\(\displaystyle ={4\over5}\)

LESSON 2
A professor has noticed that even though attendance is not a component of the grade for his class, students who attend regularly obtain better grades. In fact, the probability of attending regularly and receiving an “A” is 0.4, while the probability of not attending regularly and receiving an “A” is 0.1. The probability of attending class regularly is 0.7. Find the probability of
(a) a student receiving an “A” given that he/she attends class regularly;
(b) a student receiving an “A” given that he/she does not attend class regularly.
SOLUTION
(a) P(A│regular attendance)
\(=\displaystyle {P(A ∩ regular\: attendance)\over P(regular \:attendance)}\)
\(\displaystyle ={0.4\over0.7}\)
\(\displaystyle ={4\over7}\)

(b) P(infrequent attendance)
\(=1-P(regular \:attendance)\)
\(\displaystyle =1-0.7\)
\(=0.3\)
\(P(A \:|\:infrequent \:attendance)\)
\(\displaystyle {P(A∩infrequent \:attendance)\over P(infrequent \:attendance)}\)
\(\displaystyle {0.1\over0.3}={1\over3}\)

LESSON 3

The table shows the colour of hair and the colour of eyes of a sample of 750 people from a particular population.

Colour of hair

Black

Dark

Medium

Fair

Auburn

Total

Colour of eyes

Blue

5

50

67

65

23

210

Brown

15

93

98

91

48

345

Green

0

37

55

64

39

195

Total

20

180

220

220

110

750

Calculate, the probability that a person, selected at random from this sample, has:

(i) fair hair;

(ii) auburn hair and blue eyes;

(iii) either auburn hair or blue eyes but not both;

(iv) green eyes, given that the person has fair hair;

(v) fair hair, given that the person had green eyes.

SOLUTION

(i) \(\displaystyle P(fair\: hair)={220\over750}={22\over75}\)
(ii) \(P(auburn \:hair \:and \:blue\: eyes)={23\over750}\)
(iii) \(\displaystyle P(auburn \:or \:blue\: eyes \:but \:not \:both)={5+50+67+65+48+39 \over 750}={274\over750}={137\over375}\)
(iv) \(\displaystyle P(G│F)={64\over220}={16\over55}\)
(v) \(\displaystyle P(F│G)={64\over195}\)


INDEPENDENT EVENTS

Two events are independent if the probability of the occurrence of one event does not affect the occurrence of the other event. That is to say that \(P(A│B)=P(A)\) and \(P(B│A)=P(B)\).
Furthermore,
\(\displaystyle P(A│B)={P(A∩B)\over P(B)}\)
\(\displaystyle ⟹P(A)={P(A∩B)\over P(B)}\)
\(⟹P(A)×P(B)=P(A∩B)\)
Recall: \(P(A∪B)=P(A)+P(B)-P(A∩B)\)
Therefore, for independent events,
\(P(A∪B)=P(A)+P(B)-P(A)P(B)\)
This probability is usually associated with the words “and” or “and then.”

LESSON 1

The events \(A\) and \(B\) are such that \(P(A)=0.5, P(B)=0.3\).
(a) Evaluate \(P(A∪B)\) when
(i) \(A, B\) are mutually exclusive,
(ii) \(A, B\) are independent.

(b) Given that \(P(A∪B)=0.7\), find the value of \(P(B│A)\).

SOLUTION

(a) (i) \(P(A∪B)=P(A)+P(B)\)
\(=0.5+0.3\)
\(=0.8\)
(ii) \(P(A∩B)=P(A)×P(B)\)
\(=0.5×0.3\)
\(=0.15\)
\(P(A∪B)=P(A)+P(B)-P(A∩B)\)
\(=0.5+0.3-0.15\)
\(=0.65\)
(b) \(P(A∪B)=P(A)+P(B)-P(A∩B)\)
\(0.7=0.5+0.3-P(A∩B)\)
\(P(A∩B)=0.8-0.7\)
\(=0.1\)
\(\displaystyle P(B│A)={0.1\over0.5}\)
\(=0.2\)

LESSON 2
Each weekday, Maria travels to school by bus. Sometimes she arrives late.
  • \(L\) is the event that Maria arrives late.
  • \(R\) is the event that it is raining
You are given that \(P(L)=0.15, P(R)=0.22\) and \(P(L│R)=0.45\)
(i) Use this information to show that the events \(L\) and \(R\) are not independent.
(ii) Find \(P(L∩R)\).

(iii) Draw a Venn diagram showing the events \(L\) and \(R\), and fill in the probability corresponding to each of the four regions of your diagram.

SOLUTION

(i) If \(L\) and \(R\) are independent then
\(P(L│R)=P(L)\)
\(0.45≠0.15\)
Therefore, \(L\) and \(R\) are not independent.
(ii) \(\displaystyle P(L│R)={P(L∩R)\over P(R)}\)
\(P(L│R)×P(R)=P(L∩R)\)
\(0.45×0.22=P(L∩R)\)
\(0.099=P(L∩R)\)

(iii)