2.2 laws of probability (1)

2.2 Laws of probability - Targets1. Calculate conditional probability from a table.2. Apply the law P(A B) = P(A) + P(B) – P(A ∩ B) to all ∪

events. 3. Apply the law P(A ∩ B) = P(A)P(B I A) to events which

are not independent. 4. Solve probability problems using the laws of

probability or tree diagrams.

Stretch and challenge:5. How does a tree diagram with 2 outcomes on

each branch link in with the binomial expansion/probability?

Conditional probability• A room contains 25 people. The table below shows

the numbers of each sex and whether or not they are wearing glasses.

• A person is selected at random.

• F is the event that the person selected is female.

• G is the event that the person selected is wearing glasses.

Conditional probability

• There are a total of 9 people wearing glasses so P(G) = .

• Only 4 of the nine males are wearing glasses so the probability of a male wearing glasses is while for the females the probability is .

• This means the probability of event G occurring is affected by whether or not event F has occurred.

• The 2 events are not independent.

Conditional probability• The conditional probability that the

person selected is wearing glasses given that they are female is denoted P(G I F).

P(A I B) denotes the probability that event A happens given that event B happens.

Two events A and B are independent if: P(A) = P(A I B)

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:a) P(M)There are 40 male students out of a total of 70 students. = 0.571

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:b) P(R)21 students are studying Russian.P(R) = = 0.3

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:c) P(M I R)There are 21 students studying Russian and 14 are male. = 0.67

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:d) P(M ∩ R)There are 14 students who are both male and studying Russian = 0.2

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:e) P(M ∪ R)There are 17 + 9 + 14 + 7 = 47 students who are either male or studying Russian (or both). = 0.671

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:f) P(R I M)There are 40 male students of which 14 are studying Russian = 0.35

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:g) P(M’)There are 30 students who are not male = 0.429

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:h) P(R’)There are 29 + 20 = 49 students not studying Russian = 0.7

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:i) P(R I M’)Of the 30 who are not male, 7 are studying Russian. = 0.233

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:j) P(R’ I M)Of the 40 males, 17 + 9 = 26 are not studying Russian = 0.65

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:k) P(M’ ∩ R)There are 7 students who are not male and are studying Russian, = 0.1

Example 1 – 3.5 Page 44/45Students on the first year of a science course at a university take an optional language module. The number of students of each sex choosing each available language is shown below.

A student is selected at randomM is the probability that the student selected is maleR is the event that the student selected is studying RussianWrite down the value of:l) P(M ∪ R’)There are 40 + 12 + 11 = 63 students who are either male or not studying Russian (or both), = 0.9.

Addition law of probability• In 2.1 we saw that if A and B are mutually

exclusive events then P(A B) = P(A) + P(B).∪

• A more general form of this expression which applies whether or not A and B are mutually exclusive events is

P(A B) = P(A) + P(B) - P(A ∩ B) ∪

This is known as the addition law of probability

Addition law of probability

• Diagram on page 45 and subsequent working on 45/46 illustrates this

• Also see next example for how a numerical example proof

Example 2150 students at a catering college have to choose between 3 tasks as part of their final assessment. The table on example 3.6 on page 46 gives their choices.A student is randomly selected.C denotes the event that the selected student chooses cake baking.D denotes the event that the selected student is female. Verify that P(C D) = P(C) + P(D) – P(C ∩ D)∪

Example 2P(C D) = P(C) + P(D) – P(C ∩ D)∪P(C) => There are 54 + 26 = 80 cake-baking students.P(C) = P(D) => There are 26 + 18 + 16 = 60 female studentsP(D) = P(C ∩ D) => There are 26 female cake-baking students.P(C ∩ D) = P(C) + P(D) – P(C ∩ D) = + - P(C) + P(D) – P(C ∩ D) = There are 54 + 26 + 18 + 16 = 114 students who are either cake-baking or female or bothP(C D) = = P(C) + P(D) – P(C ∩ D) as required. ∪

Task 1

• Exercise 3E

• Questions 1 and 3.

Task 1 Answers1a 1b 1c 1d 1e 1f 1g 1h 1i 1j

2a 2b 2c 2d 2e 2f 2g 2h 2i

3a 3b 3c 3d 3e 3f 3g 3h 3i 03j A and C not independent

Multiplication law• If A and B are independent events, then P(A

∩ B) = P(A)P(B).

• This is a special case of the multiplication law for probability.

P(A ∩ B) = P(A)P(B I A)

Multiplication law• Verify this by example

• A person is selected at random• F is the event that the person is selected is female.• G is the event that the person selected is wearing glasses.

• P(F ∩ G), the probability that the person selected is a female wearing glasses is or 0.2.

• P(F) = = 0.64 and P(G I F) = = 0.3125

• P(F)P(G I F) = 0.2 • P(F)P(G I F) = P(F ∩ G)

Example 3 – 3.7 Page 49Sheena buys ten apparently identical oranges.Unknown to her, two of the oranges are rotten.She selects two of the ten oranges at random and gives them to her grandson. Find the probability that:a) Both the oranges are rottenb) Exactly one of the oranges is rotten.

Example 4 – 3.8 Page 50When Bali is on holiday she intends to go for a five-mile run before breakfast each day. However, sometimes she stays in bed instead.The probability that she will go for a run on the first morning is 0.7.Thereafter, the probability she will go for a run is 0.7 IF she went on the previous morning and 0.6 if she did not. Find the probability that on the first 3 days of the holiday she will go for:a) 3 runsb) Exactly 2 runs

Task 2

• Exercise 3F

• Pages 50-52

• Questions 1, 3, 5 and 6

Task 2 Answers1a 0.1951b 0.499

3ai 0.01523aii 0.1823aiii 0.227

3bi 0.09093bii 0.1363biii 0.4093biv 0.218

5a 0.1965b 0.02405c 0.08405d 0.09605e 0.2405f 0.2285g 0.192

6a 0.04616b 0.2336c 0.01216d 0.2796e 0.01016f 0.2666g 0.152

2.2 Laws of probability – Key Points• P(A I B) denotes the probability that event A

happens given that event B happens.

• Two events A and B are independent if: P(A) = P(A I B)

• Addition law of probability – P(A B) = P(A) + P(B) - P(A ∩ B) ∪

• Multiplication law of probability– P(A ∩ B) = P(A)P(B I A)