Statistics

Probability: Venn Diagrams, Independent & Mutually Exclusive Events, and Conditional Probability

Year 12 · Year 13

✓By the end of this lesson students will be able to represent events and probabilities using Venn diagrams.
✓By the end of this lesson students will be able to distinguish between independent and mutually exclusive events.
✓By the end of this lesson students will be able to calculate probabilities for independent and mutually exclusive events using appropriate formulae.
✓By the end of this lesson students will be able to understand and apply the concept of conditional probability.
✓By the end of this lesson students will be able to solve complex probability problems involving combinations of these concepts.

Key concepts

Venn Diagrams

Venn diagrams are visual representations used to show the relationships between different sets or events within a sample space. The universal set (ξ) represents all possible outcomes. Circles within the rectangle represent specific events. The intersection (A ∩ B) represents outcomes common to both events A and B. The union (A U B) represents outcomes in event A, event B, or both. The complement (A') represents outcomes not in event A.

P(A U B) = P(A) + P(B) - P(A ∩ B) P(A') = 1 - P(A)

Mutually Exclusive Events

Two events, A and B, are mutually exclusive if they cannot both occur at the same time. This means they have no outcomes in common. In a Venn diagram, their circles would not overlap. If A and B are mutually exclusive, the probability of their intersection is 0.

If A and B are mutually exclusive, then P(A ∩ B) = 0. Consequently, P(A U B) = P(A) + P(B).

Independent Events

Two events, A and B, are independent if the occurrence of one does not affect the probability of the other occurring. For example, rolling a 6 on a die and then flipping a head on a coin are independent events. To test for independence, we check if the probability of their intersection is equal to the product of their individual probabilities.

If A and B are independent, then P(A ∩ B) = P(A)P(B).

Conditional Probability

Conditional probability is the probability of an event occurring, given that another event has already occurred. It effectively reduces the sample space to the outcomes of the event that is known to have occurred. The notation P(A|B) means 'the probability of event A occurring given that event B has occurred'.

P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0. Rearranging, P(A ∩ B) = P(A|B)P(B). Similarly, P(B|A) = P(A ∩ B) / P(A), provided P(A) > 0.

Key facts to remember

1The sum of probabilities of all possible outcomes in a sample space is 1.
2P(A') = 1 - P(A) (Complement Rule).
3P(A U B) = P(A) + P(B) - P(A ∩ B) (General Addition Rule).
4If A and B are mutually exclusive, P(A ∩ B) = 0, so P(A U B) = P(A) + P(B).
5If A and B are independent, P(A ∩ B) = P(A)P(B) (Multiplication Rule for Independent Events).
6Conditional probability: P(A|B) = P(A ∩ B) / P(B).
7Rearrangement of conditional probability: P(A ∩ B) = P(A|B)P(B).
8Events are mutually exclusive if they cannot happen at the same time; events are independent if the occurrence of one does not affect the probability of the other.

Worked examples

Example 1

In a class of 30 students, 18 study History (H), 15 study Geography (G), and 7 study neither. A student is chosen at random.

IFirst, determine the number of students who study both History and Geography. Let N(H) be the number studying History, N(G) for Geography, N(ξ) for the total class, and N(H U G)' for neither.

IIN(H U G) = N(ξ) - N(H U G)' = 30 - 7 = 23.

IIIUsing the formula N(H U G) = N(H) + N(G) - N(H ∩ G):

IV23 = 18 + 15 - N(H ∩ G)

V23 = 33 - N(H ∩ G)

VIN(H ∩ G) = 33 - 23 = 10. So, 10 students study both.

VIINow, draw a Venn diagram and fill in the regions:

VIIIOnly History: N(H) - N(H ∩ G) = 18 - 10 = 8

9Only Geography: N(G) - N(H ∩ G) = 15 - 10 = 5

10Both: 10

11Neither: 7

12Check total: 8 + 5 + 10 + 7 = 30.

13a) Find the probability that the student studies History but not Geography.

14P(H ∩ G') = N(H ∩ G') / N(ξ) = 8 / 30 = 4/15.

15b) Find the probability that the student studies Geography given that they study History.

16This is P(G|H). Using the formula P(G|H) = P(G ∩ H) / P(H).

17P(G ∩ H) = 10 / 30 = 1/3.

18P(H) = 18 / 30 = 3/5.

19P(G|H) = (1/3) / (3/5) = 1/3 × 5/3 = 5/9.

20c) Determine if studying History and studying Geography are independent events.

21For independence, P(H ∩ G) must equal P(H)P(G).

22P(H ∩ G) = 10 / 30 = 1/3.

23P(H) = 18 / 30 = 3/5.

24P(G) = 15 / 30 = 1/2.

25P(H)P(G) = (3/5) × (1/2) = 3/10.

26Since P(H ∩ G) = 1/3 and P(H)P(G) = 3/10, and 1/3 ≠ 3/10, the events are not independent.

Answer

a) P(H ∩ G') = 4/15 b) P(G|H) = 5/9 c) The events are not independent.

Always draw a Venn diagram first for problems involving multiple events and their overlaps, as it helps to visualise the numbers in each region.

Example 2

Events A and B are such that P(A) = 0.6, P(B) = 0.3, and P(A U B) = 0.72.

Ia) Find P(A ∩ B).

IIUsing the addition rule: P(A U B) = P(A) + P(B) - P(A ∩ B).

III0.72 = 0.6 + 0.3 - P(A ∩ B)

IV0.72 = 0.9 - P(A ∩ B)

VP(A ∩ B) = 0.9 - 0.72 = 0.18.

VIb) Determine if A and B are mutually exclusive.

VIIFor mutually exclusive events, P(A ∩ B) must be 0.

VIIISince P(A ∩ B) = 0.18 ≠ 0, A and B are not mutually exclusive.

9c) Determine if A and B are independent.

10For independent events, P(A ∩ B) must equal P(A)P(B).

11P(A)P(B) = 0.6 × 0.3 = 0.18.

12Since P(A ∩ B) = 0.18 and P(A)P(B) = 0.18, the events A and B are independent.

Answer

a) P(A ∩ B) = 0.18 b) A and B are not mutually exclusive. c) A and B are independent.

Mutually exclusive and independent are distinct concepts. Events cannot be both mutually exclusive and independent unless one event has a probability of 0.

Example 3

A factory produces items on two machines, M1 and M2. Machine M1 produces 60% of the items and M2 produces 40%. 5% of items from M1 are defective, and 3% of items from M2 are defective. An item is chosen at random.

ILet M1 be the event that an item is from Machine M1, and M2 be from Machine M2.

IILet D be the event that an item is defective.

IIIWe are given: P(M1) = 0.6, P(M2) = 0.4.

IVP(D|M1) = 0.05 (probability of defective given it's from M1).

VP(D|M2) = 0.03 (probability of defective given it's from M2).

VIa) Find the probability that an item is defective.

VIIWe need P(D). We can use the law of total probability:

VIIIP(D) = P(D ∩ M1) + P(D ∩ M2).

9Using P(A ∩ B) = P(A|B)P(B):

10P(D ∩ M1) = P(D|M1)P(M1) = 0.05 × 0.6 = 0.03.

11P(D ∩ M2) = P(D|M2)P(M2) = 0.03 × 0.4 = 0.012.

12P(D) = 0.03 + 0.012 = 0.042.

13b) Given that an item is defective, find the probability that it came from Machine M1.

14This is P(M1|D). Using the conditional probability formula:

15P(M1|D) = P(M1 ∩ D) / P(D).

16We already found P(M1 ∩ D) = 0.03 and P(D) = 0.042.

17P(M1|D) = 0.03 / 0.042.

18P(M1|D) = 30 / 42 = 5 / 7.

Answer

a) P(D) = 0.042 b) P(M1|D) = 5/7

This type of problem, often called Bayes' Theorem in more advanced contexts, is a common application of conditional probability. Drawing a probability tree diagram can be very helpful for visualising these steps.

Common mistakes

✗Confusing mutually exclusive events with independent events. They are distinct concepts; if events are mutually exclusive, they cannot be independent (unless one has zero probability).
✗Incorrectly applying the addition rule P(A U B) = P(A) + P(B) when events are not mutually exclusive (forgetting to subtract P(A ∩ B)).
✗Incorrectly applying the multiplication rule P(A ∩ B) = P(A)P(B) when events are not independent.
✗Misinterpreting the order in conditional probability, e.g., confusing P(A|B) with P(B|A).
✗Errors in populating Venn diagrams, especially with 'only A' vs 'A' or 'neither' regions.

Exam tips

★Always draw a Venn diagram for problems involving two or three events and their overlaps; it helps to organise the given information and calculate unknown probabilities.
★Clearly state your assumptions, especially when assuming independence or mutual exclusivity, or when testing for them.
★Before applying any formula, check if the events are mutually exclusive or independent. This will guide you to the correct formula.
★Write down all known probabilities and what you need to find. This structured approach helps in breaking down complex problems.

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