Probability Diagrams (AQA GCSE Statistics: Higher): Revision Note

• A questions may not say "by drawing a sample space diagram", so you may have to decide to do it on your own

Two fair six-sided dice are rolled.

(a) Find the probability that the sum of the numbers showing on the two dice is an odd number greater than 5, giving your answer as a fraction in simplest form.

Draw a sample space diagram (in this case a grid) to show all the possible outcomes

Circle the possibilities that are odd numbers greater than 5.
(5 is not included!)

Count the number of possibilities that are circled (12) and divide that by the total number of possibilities in the diagram (36)

$\frac{12}{36}$

Finally, simplify your answer

$\frac{12}{36}=\frac{12\times 1}{12\times 3}=\frac{1}{3}$

$\frac{\mathbf{1}}{\mathbf{3}}$

(b) Given that the sum of the numbers showing on the two dice is an odd number greater than 5, find the probability that one of the dice shows the number 2.  Give your answer as a fraction in its simplest form.

From part (a) you already know there are 12 ways to get an odd number greater than 5

Now find how many of those ways (i.e., how many of the circled possibilities in the sample space grid) have one of the dice showing the number 2
There are two of these: (2, 5) and (5, 2)
So the probability we are looking for is 2 divided by 12

$\frac{2}{12}$

Finally, simplify your answer

$\frac{2}{12}=\frac{2\times 1}{2\times 6}=\frac{1}{6}$

$\frac{\mathbf{1}}{\mathbf{6}}$

• If you had to complete a two-way table (or add the 'total' rows and columns), double-check that your numbers are correct

  • If there are errors in the table, probabilities calculated from it will be incorrect and you could lose marks

At an art group, children are allowed to choose between colouring, painting, clay modelling and sketching.

A total of 60 children attend and are split into two classes: class A and class B.

The data for the art group is represented in the following two-way table.

A child is selected at random from the art group. Find the probability that the child

(a) chose colouring

A total of 13 children chose colouring, out of 60 students in total

$\frac{13}{60}$

(b) is a member of class B who chose sketching

4 children are in class B and chose sketching
We are still selecting out of 60 children in total

$\frac{4}{60}$

The simplified answer $\frac{1}{15}$ is also accepted, but is not necessary to get full marks

(c) chose painting, given that the child is a member of class A

Now the choice is only 'out of' the 30 children in class A
8 children in that class chose painting

$\frac{8}{30}$

The simplified answer $\frac{4}{15}$ is also accepted, but is not necessary to get full marks

(d) is a member of class B, given that the child chose sketching.

Note that this is not the same thing that is asked in part (b)!

The choice here is out of the 12 students who chose sketching
4 of those children are in class B

$\frac{4}{12}$

The simplified answer $\frac{1}{3}$ is also accepted, but is not necessary to get full marks

• If you had to create the Venn diagram (or fill in some of the numbers), double-check that your numbers are correct

  • If there are errors in the Venn diagram, probabilities calculated from it will be incorrect and you could lose marks

In a class of 30 students, students can study Spanish, German, both, or neither.

The data for the class is represented on the following Venn diagram.

A student is chosen from the class at random. Find the probability that the student

(a) studies German

11 (3+8) students study Spanish, out of 30 students in total

$\frac{11}{30}$

(b) studies Spanish but not German

12 students study Spanish but not German (the part of the 'S' oval that's outside the 'G' oval)
We are still selecting out of 30 students in total

$\frac{12}{30}$

The simplified answer $\frac{2}{5}$ is also accepted, but is not necessary to get full marks

(c) studies German, given that they study Spanish

Now the choice is out of the 15 (12+3) students who study Spanish
3 of those students also study German

$\frac{3}{15}$

The simplified answer $\frac{1}{5}$ is also accepted, but is not necessary to get full marks

(d) studies Spanish, given that they do not study German .

Note that this is not the same thing that is asked in part (b)!

There are 19 (12+7) students who do not study German (all the numbers outside the 'G' oval)
12 of those students study Spanish

$\frac{12}{19}$

The following Venn diagram shows the probabilities associated with two events, $A$ and $B$.

It is known that  $\mathrm{P}\left(A\mathrm{or}B\right)=0.32$.

(a) Find the values of $x$ and $y$.

To find $x$we can either use the general addition law $\mathrm{P}\left(A\mathrm{or}B\right)=\mathrm{P}\left(A\right)+\mathrm{P}\left(B\right)-\mathrm{P}\left(A\mathrm{and}B\right)$

$\begin{array}{rcl}0.32& =& \left(x+0.03\right)+\left(0.03+0.12\right)-0.03\\ 0.32& =& x+0.15\\ 0.17& =& x\end{array}$

Or you could just notice that $\mathrm{P}\left(A\mathrm{or}B\right)$ has to be the sum of $x$, 0.03 and 0.12

To find the value of $y$, remember that these are probabilities
So all of them must add up to 1

$\begin{array}{rcl}0.17+0.03+0.12+y& =& 1\\ y+0.32& =& 1\\ y& =& 1-0.32\\ y& =& 0.68\end{array}$

$x=0.17,y=0.68$

(b) Show that $A$ and $B$ are independent.

Remember, if $\mathrm{P}\left(A\mathrm{and}B\right)=\mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right)$ is true, then A and B are independent

$\begin{array}{l}\mathrm{P}\left(A\mathrm{and}B\right)=0.03\\ \\ \mathrm{P}\left(A\right)=0.17+0.03=0.2\mathrm{P}\left(B\right)=0.03+0.12=0.15\\ \\ \mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right)=0.2\times 0.15=0.03\end{array}$

$\mathrm{P}\left(A\mathrm{and}B\right)=\mathrm{P}\left(A\right)\times \mathrm{P}\left(B\right)$, so A and B are independent

(c) Find the probability of $B$ given $A$.

Use  $\mathrm{P}\left(B|A\right)=\frac{\mathrm{P}\left(A\mathrm{and}B\right)}{\mathrm{P}\left(A\right)}$

$P\left(B|A\right)=\frac{0.03}{0.2}=0.15$

(You could also use the fact that because $A$ and $B$ are independent,  $\mathrm{P}\left(B|A\right)=\mathrm{P}\left(B\right)$ )

$\mathrm{P}\left(B|A\right)=0.15$

• Remember  $\mathrm{P}\left(\mathrm{not}A\mathit{\right)}\mathit{=}1\mathit{-}P\mathit{\left(}A\mathit{\right)}$

• Tree diagrams have built-in checks

  • the probabilities for each pair of branches should add up to 1

  • the probabilities for all final outcomes should add up to 1

• When multiplying along branches with fractions, don't cancel fractions in your working - having the same denominator makes them easier to add together!

Lisa drives through two sets of traffic lights on her way to work.

Each set of traffic lights has only two options: green or red.

The probability of the first set of traffic lights being on green is $\frac{5}{7}$.

The probability of the second set of traffic lights being on green is $\frac{8}{9}$.

(a) Draw and label a tree diagram. Be sure to show the probabilities of every possible outcome.

Work out the probabilities of each set of traffic lights being on red, R
Use P(red) = 1 - P(green)

$\mathrm{P}\left(1^{\mathrm{st}}\mathrm{R}\right)=1-\mathrm{P}\left(1^{\mathrm{st}}\mathrm{G}\right)=1-\frac{5}{7}=\frac{2}{7}$
$\mathrm{P}\left(2^{\mathrm{nd}}\mathrm{R}\right)=1-\mathrm{P}\left(2^{\mathrm{nd}}\mathrm{G}\right)=1-\frac{8}{9}=\frac{1}{9}$

Draw the branches (with a label of G or R on the ends)
Write the probabilities above each branch
Calculate probabilities of each outcome by multiplying along the branches from left to right

(b) Find the probability that both sets of traffic lights are on red.

This is the answer for P(R, R) from the tree diagram

$\frac{\mathbf{2}}{\mathbf{63}}$

(c) Find the probability that at least one set of traffic lights is on red.

This means the 1st is green and the 2nd is red
Or the 1st is red and the 2nd is green
Or the 1st is red and the 2nd is red ('at least one' could mean both)
Add the probabilities for the separate cases

$\mathrm{P}\left(\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{R}\right)=\mathrm{P}\left(\mathrm{G},\mathrm{R}\right)+\mathrm{P}\left(\mathrm{R},\mathrm{G}\right)+\mathrm{P}\left(\mathrm{R},\mathrm{R}\right)=\frac{5}{63}+\frac{16}{63}+\frac{2}{63}=\frac{23}{63}$

$\frac{\mathbf{23}}{\mathbf{63}}$

Alternative Method
At least one red means all the possible cases shown except two greens
So P(at least 1 red) = 1 - P(two greens)

$1-\mathrm{P}\left(\mathrm{G},\mathrm{G}\right)=1-\frac{40}{63}=\frac{23}{63}$ 

$\frac{\mathbf{23}}{\mathbf{63}}$

Liana has 10 pets: 7 guinea pigs (G) and 3 rabbits (R).

Liana is choosing two pets to feature in her latest online video. 

Firstly, she is going to choose one of the pets at random. 

Once she has carried that pet to her video studio, she is going to go back and choose a second pet at random, to also feature in the video.

(a) Draw and label a tree diagram including the probabilities of all possible outcomes.

For the 1st pet chosen, there will be a 7/10 probability of choosing a guinea pig, and a 3/10 probability of choosing a rabbit

If the first pet is a guinea pig, there will only be 6 guinea pigs and 3 rabbits left (9 animals total)
So for the second pet the probability of choosing a guinea pig would be 6/9, and probability of choosing a rabbit would be 3/9

If the first pet is a rabbit, there will only be 7 guinea pigs and 2 rabbits left (9 animals total)
So for the second pet the probability of choosing a guinea pig would be 7/9, and probability of choosing a rabbit would be 2/9

Put these probabilities into the correct places on the tree diagram, and then multiply along the branches to find the probabilities for each outcome

(b) Write down the probability that the second pet is a guinea pig, given that the first pet was a rabbit.

This conditional probability can be done by looking at the branches, without needing the conditional probability formula

This is the probability on the second pet 'G' branch that follows 'R' for the first pet

$\frac{7}{9}$

Using the conditional probability formula would also give this answer

(c) Find the probability that Liana chooses two rabbits.

As we have already calculated this probability in the tree diagram, we can just write the answer down

$\mathrm{P}\left(\mathrm{two}\mathrm{rabbits}\right)=\mathrm{P}(\mathrm{R},\mathrm{R})=\frac{6}{90}$

The simplified answer $\frac{1}{15}$ is also accepted, but is not necessary to get full marks

(d) Find the probability that Liana chooses two different kinds of animal.

This would be "G and R" or "R and G" so we need to add two of the final probabilities

$\mathrm{P}\left(\mathrm{two}\mathrm{different}\mathrm{kinds}\right)=\mathrm{P}\left(\mathrm{G},\mathrm{R}\right)+\mathrm{P}\left(\mathrm{R},\mathrm{G}\right)=\frac{21}{90}+\frac{21}{90}=\frac{42}{90}$

$\mathrm{P}\left(\mathrm{two}\mathrm{different}\mathrm{kinds}\right)=\frac{42}{90}$ 

The simplified answer $\frac{7}{15}$ is also accepted, but is not necessary to get full marks

You could also do this question using 1 - P(both the same)