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Probability theory is the branch of Mathematics concerned with analysis of random phenomena. The central objects of probability theory are random variables and events. Although an individual coin toss or roll of a die is a random event, if repeated many times the sequence of random events will exhibit certain statistical patterns, which can be studied and predicted. Two representative Mathematics results describing such patterns are the law of large numbers.As a Mathematical foundation for statistics, probability theory is essential to many human activities that involve quantitative analysis of large sets of data. Methods of probability theory also apply to description of complex system given only partial knowledge of their state, as in statistical mechanics.The Mathematics theory of probability has its roots in attempts to analyze games of change by Gerolamo Cardano in the 16th Century.

Initially, probability theory mainly considered discrete events, and its methods were mainly combinatorial. Eventually, analytical consideration compelled the incorporation of continuous variables into the theory. This culminated in modern probability theory, the foundation of which was variables into the theory.Theoretical ProbabilityWhenever an event is carried out like tossing a coin, there is always an outcome. For example when a coin is tossed once, either a side with “Tail” or “Head” appears on top. Each of these two possibilities is called an outcome when a die is rolled once, there are six possible outcomes, a score of {1, 2, 3, 4, 5, 6}. If these scores have equal chances of appearing, then the die is said to be fair.Probability (P) is defined as, ‘the number of ways the events can occur’ i.e. the number of possible outcomes

For example: A fair die is rolled once. Calculate probability of getting

0 ≤ P(A) ≥1. 0 ≤ P(A)≤1The probability of an event equals one if the occurancy of that event is “certain”. For example, if we roll a die what is the probability of getting a score less than 7? There are 6 possible outcomes: {1, 2, 3, 4, 5, 6}.All these outcomes are scores less than 7 so P(a score less than 7) = 6/6 = 1. What is P (a score of 7)?No outcome gives a score of 7, so P (7) = 0.The probability of a certain event is 1.The probability of an impossible event is 0.Exercise

9/20.

Note:In order to get the scores of each team that played on that week end, you need to watch the matches or get the results from other sources e.g. newspapers, football fans, electronic media etc.In theory, when a fair coin is tossed twice, we expect to get one head or one tail, if it is tossed 10 times then results would be 5 heads or 5 tails.

If the tossing is done practically / experimentally the results may not be the same as those obtained theoretically. So if a coin is tossed 50 times, the number of times a head or tail appears may not be the same. For instance you may get 29 heads and 21 tails but if the coin is tossed several times like 100 times, the number of times the head appears is approximately equal to the number of times the tail appears.Insert an experiment of determining the number of times a head appears when a coin is tossed

(i) more than 169 cm tall?

(ii) less than 160 cm tall?b) The basket ball coach wants to pick a team. He says that all players must be at least 160 cm tall. What is the probability that a student chosen at random will meet the coach’s requirements?[/td][td] [/td][/tr][/table] Possibility space in Cartesian DiagramsAll the possible outcomes when a coin is tossed or a die is rolled or when a coin and a die are tossed form what is called possibility space.

For example:

a) When a coin is tossed once the possibility space is {H, T}

b) A die is rolled, the possibility space is {1, 2, 3, 4, 5, 6}

c) 2 dice are rolled, the possibility space is (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 3), (6, 5), (6, 6)

d) A die and a coin tossed, the possibility space is (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6) (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6).

Note:

Use the possibility space for throwing two dice to calculate these probabilities

For example:Mutually Excusive Vs Independent

It is not uncommon for people to confuse the concepts of mutually exclusive events and independent events.

Definition of a mutually exclusive event

If event A happens, then event B cannot, or vice-versa. The two events “it rained on Tuesday” and “It did not rain on Tuesday” are mutually exclusive events. When calculating the probabilities for exclusive events you add the probabilities

.Independent events

The outcome of event A, has no effect on the outcome of event B. Such as “It rained on Tuesday” and “My chair broke at work”. When calculating the probabilities for independent events you multiply the probabilities. You are effectively saying what is the chance of both events happening bearing in mind that the two were unrelated.To be or not be …?So, if A and B are mutually exclusive, they cannot be independent. If A and B are independent, they cannot be mutually exclusive. However, if the events were “it rained today” and “I left my umbrella at home” they are not mutually exclusive, but they are probably not independent either, because one would think that you would be less likely to leave your umbrella at home on days when it rains. That fact aside use the following to understand the definition.

Example of a mutually exclusive event

What happens if we want to throw 1 and 6 in any order? This now means that we do not mind if the first die is either 1 or 6, as we are still in with a chance. But with the first die, if 1 falls uppermost, clearly it rules out the possibility of 6 being uppermost, so the two outcomes, 1 and 6 are exclusive. One result directly affects the other. In this case, the probability of throwing 1 or 6 with the first die is the sum of the two probabilities, 1/6 + 1/6 = ⅓.The probability of the second die being favourable is still 1/6 as the second die can only be one specific number, a 6 if the first die is 1, and vice versa.Therefore, the probability of throwing 1 and 6 in any order with two dice is ⅓ x 1/6 = 1/18. Note that we multiplied the last two probabilities as they were independent of each other!!!

Example of an independent event

The probability of throwing a double three with two dice is the result of throwing three with the first die and three with the second die. The total possibilities are, one from six outcomes for the first event and one from six outcomes for the second, therefore (1/6) * (1/6) = 1/36th or 2.77%.The two events are independent, since whatever happens to the first die cannot affect the throw of the second, the probabilities are therefore multiplied, and remain 1/36th.The probability of getting H and H when 2 coins are tossed {HT, HH, TH, TT} is ¼.Note that P (H, H) = ¼ = ½ x ½ = P (H) x P (H)By definition if A and B are independent events P ( (A, B) = P (A) x P (B)We can also determine the probability of independent events by using Tree diagram.

When two coins are tossed, the outcomes can be illustrated in the following diagram:

First Second Outcome

toss toss

So, P (H, H) = P (H) x P (H) = ½ x ½ =¼ P (H, T) = P (H) x P (T)= ½ x ½ =¼ P (T, T) = P (T) x P (T)= ½ x ½ =¼ P (T, H) = P (T) x P (H)= ½ x ½ =¼The above diagram is what is called Tree diagram. The above Tree diagram can be used to determine probabilities of events e.g.P (one head) = P (H, T) + P (T, H) = ¼ + ¼ = ½.The above probabilities have been added together because the events (H, T) and (T, H) are mutually exclusive. This means that when one event occurs the other does not.

Example 1:

A bag contains 3 red and 9 white beads, two beads are taken out of it with replacement. Draw a tree diagram and use it to find these probabilities

Possibility space = 3 + 9 = 12a) P (red) = P ® = 3/12b) P (white) = P (w) = 9/12 a) P (w, w) = P (w) x P (w) = 9/12 x 9/12 = ¾ x ¾ = 9/16b) P (w, r) = P (w, r) + P (r, w) = P (w) x P ® + ( P ® x P (w) = 9/12 x 3/12 + 3/12 x 9 /12 = ¾ x ¼ + ¼ x ¾ = 3/16 + 3/16 = 6/16

Example 2:

Use the above example without replacement.

Solution:

1[sup]st[/sup] pick, the possibility space is 3 + 9 = 12 P (w) = 9/12 P ® = 3/122nd pick, the possibility space is 12 -1 = 11 P (w) depends on whether, in first pick the bead was white. If it was white, then P (w) = 8/11 If it was not, P (w) = 9/11, likewise If it was red then P ® = 2/11 If it was not red, P ® = 3/11Diagrammatically, it will be illustrated as follows.

SoP (w, w) = P (w) x P(w) = 9/12 x 8/11 = 72/132 = 18/33P (w, r) = P (w r) + P(r w) = P (w) x P® + (P ® x P(w) = 9/12 x 3/11 + 3/12 x 9 /11 = ¾ x ¼ + ¼ x ¾ = 9/44 + 9/44 = 18/44 = 9/22

Exercise:

Venn diagrams which are usually used in set theory can also be used to solve some probability problems.

Example 1: The diagram shows how children come to school by walking (W), by bicycle (B) or by car (C).

Use the information on the Venn diagram to find the probability that a child picked at random

Possibility space = 6 + 5 + 7 + 10 + 9 + 8 + 12 = 57a) children who walk = 7 + 8 + 5 + 9 = 29P (W) = 29/57b) uses a car = 6 + 5 + 10 + 9 = 30P © = 30/57 = 10/19c) uses a Bicycle = 10 + 9 + 8 + 12 = 39P (B) = 39/57 = 13/19d) Walks only = 7P (child who walks only)= 7/57e) Uses Bicycles and Car = 10 + 9 + 8 = 27P (child who uses Bicycle and Car) = 27/57 = 9/19f) Uses all the means of travel = 9P (Child uses all the means of travel) = 9/57 = 3/19

Example 2:

In a class, 15 pupils like nature study, 13 like crafts and eight like both subjects. Six pupils do not like either subject. What is the probability that a pupil picked at random from the class does not like crafts?

Solution:

Crafts ©, Nature Study (N)

Pupils who don’t like crafts = pupils who like Nature Study only + pupils who don’t like any of the subjects = 7 + 6 = 13.Possibility space: 7 + 8 + 5 + 6 = 26P (Pupil picked at random does not like Craft) = 13/26 = ½Exercise:

Initially, probability theory mainly considered discrete events, and its methods were mainly combinatorial. Eventually, analytical consideration compelled the incorporation of continuous variables into the theory. This culminated in modern probability theory, the foundation of which was variables into the theory.Theoretical ProbabilityWhenever an event is carried out like tossing a coin, there is always an outcome. For example when a coin is tossed once, either a side with “Tail” or “Head” appears on top. Each of these two possibilities is called an outcome when a die is rolled once, there are six possible outcomes, a score of {1, 2, 3, 4, 5, 6}. If these scores have equal chances of appearing, then the die is said to be fair.Probability (P) is defined as, ‘the number of ways the events can occur’ i.e. the number of possible outcomes

For example: A fair die is rolled once. Calculate probability of getting

- an even number

- a prime or an even number

- a score of 6

- a score of 8

- scores 1, 2, 3, 4, 5, 6.

0 ≤ P(A) ≥1. 0 ≤ P(A)≤1The probability of an event equals one if the occurancy of that event is “certain”. For example, if we roll a die what is the probability of getting a score less than 7? There are 6 possible outcomes: {1, 2, 3, 4, 5, 6}.All these outcomes are scores less than 7 so P(a score less than 7) = 6/6 = 1. What is P (a score of 7)?No outcome gives a score of 7, so P (7) = 0.The probability of a certain event is 1.The probability of an impossible event is 0.Exercise

- Nine pieces of paper are put in a bag. One piece has a 1 on it, one piece has a 2 on it and so on up to 9. one piece of paper is picked at random from the bag. Find the probability of these events.|The number on the paper is ‘odd’.The number on the paper is either ‘prime’ or a ‘multiple of 2’The number is less than ‘8’ and a multiple of ‘2 and 3’.

- A number is selected at random from the sets of integers from 1 to 20 inclusive. Find the probability that the number selected is.

9/20.

Note:In order to get the scores of each team that played on that week end, you need to watch the matches or get the results from other sources e.g. newspapers, football fans, electronic media etc.In theory, when a fair coin is tossed twice, we expect to get one head or one tail, if it is tossed 10 times then results would be 5 heads or 5 tails.

If the tossing is done practically / experimentally the results may not be the same as those obtained theoretically. So if a coin is tossed 50 times, the number of times a head or tail appears may not be the same. For instance you may get 29 heads and 21 tails but if the coin is tossed several times like 100 times, the number of times the head appears is approximately equal to the number of times the tail appears.Insert an experiment of determining the number of times a head appears when a coin is tossed

- 20

- 40

- 50

- 70

- 100

- 150,

- In a traffic survey, the number of passengers (including the driver) in each car passing a school was recorded and the results were as follows:

Quote:a) exactly 4 passengers?b) more than 2 passengers?c) less than the average number of passengers?2. In a survey, the heights of students of Senior 3 were recorded and results were as follows:[table][tr][td]Height of pupils[/td][td]140-149[/td][td]150-159[/td][td]160-169[/td][td]170-179[/td][td]180-189[/td][/tr][tr][td]Number of pupils[/td][td]2[/td][td]12[/td][td]14[/td][td]21[/td][td]16[/td][/tr][/table]a) What is the probability that a student picked at random is;

(i) more than 169 cm tall?

(ii) less than 160 cm tall?b) The basket ball coach wants to pick a team. He says that all players must be at least 160 cm tall. What is the probability that a student chosen at random will meet the coach’s requirements?[/td][td] [/td][/tr][/table] Possibility space in Cartesian DiagramsAll the possible outcomes when a coin is tossed or a die is rolled or when a coin and a die are tossed form what is called possibility space.

For example:

a) When a coin is tossed once the possibility space is {H, T}

b) A die is rolled, the possibility space is {1, 2, 3, 4, 5, 6}

c) 2 dice are rolled, the possibility space is (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 3), (6, 5), (6, 6)

d) A die and a coin tossed, the possibility space is (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6) (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6).

Note:

- Explain what each of the pairs in © and (d) means e.g. (4, 2), (T, 4), (4, 2) means that the first die shows up the side marked 4 and the second dies shows 2.

- (T, 4) means that a coin is tossed first and shows the “Tail” and the die shows up 4.

- © and (d) above can be put in Cartesian diagram

Use the possibility space for throwing two dice to calculate these probabilities

- P (a sum of 10 or more)

- P (an even sun)

- P (a sum of less than 12)

- P (a sum of 13).

- P (a sum of 10 or more) = 6/36 = 1/6

- P (an even sum) = 18/36 = ½

- P (a sum of less than 12) = 35/36

- P (a sum of 13) = 0/36

- A number is selected at random from the set S = {1, 2, 3, …, 18}

- Find P (prime number).

- Find P (even number).

- Find P (multiple of 6).

- Construct the possibility space for tossing 4 coins and use it to calculate these probabilities.

- P (4 heads).

- P (less than 4 heads).

- P (2 heads).

For example:Mutually Excusive Vs Independent

It is not uncommon for people to confuse the concepts of mutually exclusive events and independent events.

Definition of a mutually exclusive event

If event A happens, then event B cannot, or vice-versa. The two events “it rained on Tuesday” and “It did not rain on Tuesday” are mutually exclusive events. When calculating the probabilities for exclusive events you add the probabilities

.Independent events

The outcome of event A, has no effect on the outcome of event B. Such as “It rained on Tuesday” and “My chair broke at work”. When calculating the probabilities for independent events you multiply the probabilities. You are effectively saying what is the chance of both events happening bearing in mind that the two were unrelated.To be or not be …?So, if A and B are mutually exclusive, they cannot be independent. If A and B are independent, they cannot be mutually exclusive. However, if the events were “it rained today” and “I left my umbrella at home” they are not mutually exclusive, but they are probably not independent either, because one would think that you would be less likely to leave your umbrella at home on days when it rains. That fact aside use the following to understand the definition.

Example of a mutually exclusive event

What happens if we want to throw 1 and 6 in any order? This now means that we do not mind if the first die is either 1 or 6, as we are still in with a chance. But with the first die, if 1 falls uppermost, clearly it rules out the possibility of 6 being uppermost, so the two outcomes, 1 and 6 are exclusive. One result directly affects the other. In this case, the probability of throwing 1 or 6 with the first die is the sum of the two probabilities, 1/6 + 1/6 = ⅓.The probability of the second die being favourable is still 1/6 as the second die can only be one specific number, a 6 if the first die is 1, and vice versa.Therefore, the probability of throwing 1 and 6 in any order with two dice is ⅓ x 1/6 = 1/18. Note that we multiplied the last two probabilities as they were independent of each other!!!

Example of an independent event

The probability of throwing a double three with two dice is the result of throwing three with the first die and three with the second die. The total possibilities are, one from six outcomes for the first event and one from six outcomes for the second, therefore (1/6) * (1/6) = 1/36th or 2.77%.The two events are independent, since whatever happens to the first die cannot affect the throw of the second, the probabilities are therefore multiplied, and remain 1/36th.The probability of getting H and H when 2 coins are tossed {HT, HH, TH, TT} is ¼.Note that P (H, H) = ¼ = ½ x ½ = P (H) x P (H)By definition if A and B are independent events P ( (A, B) = P (A) x P (B)We can also determine the probability of independent events by using Tree diagram.

When two coins are tossed, the outcomes can be illustrated in the following diagram:

First Second Outcome

toss toss

So, P (H, H) = P (H) x P (H) = ½ x ½ =¼ P (H, T) = P (H) x P (T)= ½ x ½ =¼ P (T, T) = P (T) x P (T)= ½ x ½ =¼ P (T, H) = P (T) x P (H)= ½ x ½ =¼The above diagram is what is called Tree diagram. The above Tree diagram can be used to determine probabilities of events e.g.P (one head) = P (H, T) + P (T, H) = ¼ + ¼ = ½.The above probabilities have been added together because the events (H, T) and (T, H) are mutually exclusive. This means that when one event occurs the other does not.

Example 1:

A bag contains 3 red and 9 white beads, two beads are taken out of it with replacement. Draw a tree diagram and use it to find these probabilities

- P (two white beads are picked)

- P (one white and one red bead picked).

Possibility space = 3 + 9 = 12a) P (red) = P ® = 3/12b) P (white) = P (w) = 9/12 a) P (w, w) = P (w) x P (w) = 9/12 x 9/12 = ¾ x ¾ = 9/16b) P (w, r) = P (w, r) + P (r, w) = P (w) x P ® + ( P ® x P (w) = 9/12 x 3/12 + 3/12 x 9 /12 = ¾ x ¼ + ¼ x ¾ = 3/16 + 3/16 = 6/16

Example 2:

Use the above example without replacement.

Solution:

1[sup]st[/sup] pick, the possibility space is 3 + 9 = 12 P (w) = 9/12 P ® = 3/122nd pick, the possibility space is 12 -1 = 11 P (w) depends on whether, in first pick the bead was white. If it was white, then P (w) = 8/11 If it was not, P (w) = 9/11, likewise If it was red then P ® = 2/11 If it was not red, P ® = 3/11Diagrammatically, it will be illustrated as follows.

SoP (w, w) = P (w) x P(w) = 9/12 x 8/11 = 72/132 = 18/33P (w, r) = P (w r) + P(r w) = P (w) x P® + (P ® x P(w) = 9/12 x 3/11 + 3/12 x 9 /11 = ¾ x ¼ + ¼ x ¾ = 9/44 + 9/44 = 18/44 = 9/22

Exercise:

- A basket contains 6 mangoes and 4 oranges. Two fruits are removed from it without replacement. Use tree diagrams to work out the following probabilities.P (three mangoes are removed)P (a mango and two oranges are removed)

- A bag contains 5 blue pens and 3 red pens, three pens are removed with replacement. Find the probability that;

Venn diagrams which are usually used in set theory can also be used to solve some probability problems.

Example 1: The diagram shows how children come to school by walking (W), by bicycle (B) or by car (C).

Use the information on the Venn diagram to find the probability that a child picked at random

- walks

- uses a car

- uses a bicycle

- walks only

- uses a bicycle and car

- uses all the means of travel.

Possibility space = 6 + 5 + 7 + 10 + 9 + 8 + 12 = 57a) children who walk = 7 + 8 + 5 + 9 = 29P (W) = 29/57b) uses a car = 6 + 5 + 10 + 9 = 30P © = 30/57 = 10/19c) uses a Bicycle = 10 + 9 + 8 + 12 = 39P (B) = 39/57 = 13/19d) Walks only = 7P (child who walks only)= 7/57e) Uses Bicycles and Car = 10 + 9 + 8 = 27P (child who uses Bicycle and Car) = 27/57 = 9/19f) Uses all the means of travel = 9P (Child uses all the means of travel) = 9/57 = 3/19

Example 2:

In a class, 15 pupils like nature study, 13 like crafts and eight like both subjects. Six pupils do not like either subject. What is the probability that a pupil picked at random from the class does not like crafts?

Solution:

Crafts ©, Nature Study (N)

Pupils who don’t like crafts = pupils who like Nature Study only + pupils who don’t like any of the subjects = 7 + 6 = 13.Possibility space: 7 + 8 + 5 + 6 = 26P (Pupil picked at random does not like Craft) = 13/26 = ½Exercise:

- A and B are two sets of numbers such that ∩(a) = 17, ∩(B) = 10 and ∩(AUB) = 25. Use a Venn diagram to find the probability that a number picked at random from AUB is a member of A∩B.

- In a car park, there were 22 cars. Ben noticed that 12 of them were blue. He also noticed that nine had sun roofs. Only two blue cars had sun roofs. Show this information on a Venn diagram. What is the probability that a car chosen at random had not sun roof?

- AUB consists of the whole numbers from 1 to 29, so ∩(AUB) = 29, ∩(A) = 17 and ∩(B) = 23. Find the probability that;