Probability and Venn Diagrams

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Probability and Venn Diagrams

Probability is a measure of how likely something is to happen.

It always falls between 0mathbf{0} and 1mathbf{1}, with 00 being impossible and 11 being certain.

Notation: The probability of an event AA is P(A)mathbb{P}(A).

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Calculating Probabilities

To calculate the probability of an event, the formula is:

number of desired outcomestotal number of possible outcomesdfrac{color{red}text{number of desired outcomes}}{color{blue}text{total number of possible outcomes}}

Example: What is the probability of obtaining an even number when rolling a 66-sided die?

Our possible outcomes are 1,2,3,4,5,6color{blue}1,2,3,4,5,6, of which 2,4,6color{red}2,4,6 are even. This gives 3color{red}3 desired outcomes, out of 6color{blue}6 possible outcomes, so:

P(roll an even number)=36=12mathbb{P}(text{roll an even number})=dfrac{color{red}3}{color{blue}6}=dfrac{color{red}1}{color{blue}2}

 

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AND & OR

ABAcap B means (A)mathbb(A) AND (B)mathbb(B) – that both event AA and event BB happen.

ABAcup B means (A)mathbb(A) OR (B)mathbb(B) – that event AA happens or event BB happens (or both).

The probabilities of ABAcap B and ABAcup B are related by the formula:

P(AB)=P(A)+P(B)P(AB)mathbb{P}(Acup B)=mathbb{P}(A)+mathbb{P}(B)-mathbb{P}(Acap B)

 

These probabilities can be calculated with a Venn Diagram.

Example: considering dice again, if AA is “rolls a multiple of 33” and BB is “rolls 33 or 44”, then we have:

By looking at where the circles overlap, we notice there is one value, so ABAcap B contains one value.

By looking at the circles in their entirety, we notice that there are 33 values contained within circle AA or circle BB, so ABAcup B contains three values.

Since there are six values overall, we can conclude:

P(AB)=16mathbb{P}(Acap B)=dfrac{1}{6}

P(AB)=36=12mathbb{P}(Acup B)=dfrac{3}{6}=dfrac{1}{2}

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Complement of an Event

The complement of an event is the event that it doesn’t happen, for example if AA is rolls a 33 or 44 then the complement of AA, written AA', is does not roll a 33 or 44, i.e. rolls a 11,22,55 or 66.

Since any event definitely either does or does not happen, the probability of an event and its complement must add to 1mathbf{1}.

P(A)+P(A)=1mathbb{P}(A)+mathbb{P}(A')=1

On a Venn Diagram, the complement of AA is everything not contained within the circle representing AA.

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Two-Way Tables

Two way tables model two events, AA and BB, by displaying every combination of whether or not each one happens. They can either display frequency (which should be familiar from GCSE) or probability. If it displays probability, then the numbers in it excluding row totals and column totals must add to 1mathbf{1}, and the final total in the bottom right must be 11. An example is given below.

Note: the totals are probabilities for one event, e.g. the total in the AA column is the probability of event AA. Using this, we can create the most general form of a two-way table:

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Probability and Venn Diagrams Example Questions

Question 1: Inside a bag are 33 red marbles and 88 green marbles. Cate picks one marble at random. What is the probability that she has picked a green marble?

[1 mark]

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There are 1111 marbles, representing 1111 possible outcomes, and 88 marbles that give the desired outcome, so the probability is 811dfrac{8}{11}.

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Question 2: If AA has probability 0.40.4 and BB has probability 0.30.3, what is P(AB)+P(AB)mathbb{P}(Acup B)+mathbb{P}(Acap B)?

[2 marks]

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P(AB)=P(A)+P(B)P(AB)mathbb{P}(Acup B)=mathbb{P}(A)+mathbb{P}(B)-mathbb{P}(Acap B)

 

P(A)=0.4mathbb{P}(A)=0.4

 

P(B)=0.3mathbb{P}(B)=0.3

 

P(AB)=0.4+0.3P(AB)=0.7P(AB)begin{aligned}mathbb{P}(Acup B)&=0.4+0.3-mathbb{P}(Acap B)[1.2em]&=0.7-mathbb{P}(Acap B)end{aligned}

 

P(AB)+P(AB)=0.7mathbb{P}(Acup B)+mathbb{P}(Acap B)=0.7

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Question 3: Consider flipping a coin three times. Event AA is that the second flip is tails. Event BB is that there are at least two heads.

a) Represent the events AA and BB on a Venn Diagram.

b) Find P(A)mathbb{P}(A)

c) Find P(B)mathbb{P}(B)

d) Find P(AB)mathbb{P}(Acap B)

e) Find P(AB)mathbb{P}(Acup B)

f) How many possibilities do not fall within the Venn Diagram at all?

[6 marks]

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a)

 

b) 88 total events, 4 of them in circle A, so probability is 48=12dfrac{4}{8}=dfrac{1}{2}

 

c) 88 total events, 44 of them in circle B, so probability is 48=12dfrac{4}{8}=dfrac{1}{2}

 

d) This corresponds to where the circles overlap, which contains one event, so probability is 18dfrac{1}{8}

 

e) This corresponds to both circles, which contain 77 events, so probability is 78dfrac{7}{8}

 

f) 11 possibility does not lie within the Venn Diagram

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Question 4: An event has probability 0.9320.932. What is the probability of its complement?

[1 mark]

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P(complement)=1P(event)=10.932=0.068begin{aligned}mathbb{P}(text{complement})&=1-mathbb{P}(text{event})[1.2em]&=1-0.932[1.2em]&=0.068end{aligned}

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Question 5: Complete the two-way table and use it to determine P(AB)mathbb{P}(Acap B')

 

[4 marks]

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P(AB)=0.08mathbb{P}(Acap B')=0.08

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Additional Resources

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Specification Points Covered

M3 – Modelling with probability, including critiquing assumptions made and the likely effect of more realistic assumptions