Maths · 9 Probability

9.3 Probability of combined events

Before you start

The probability of a single event with equally likely outcomes: the number of outcomes you want divided by the total number of outcomes (9.1).
That probabilities run from $0$ to $1$ , and that $P (A^{'}) = 1 - P (A)$ (9.1).
Reading and filling a two-set Venn diagram, and what $A \cap B$ , $A \cup B$ and $A^{'}$ mean (1.2).
Multiplying, adding and simplifying fractions without a calculator.

Syllabus objectives covered (1)

Calculate the probability of combined events using, where appropriate: • sample space diagrams • Venn diagrams • tree diagrams.

Part 1 of 6

Combined events and sample space diagrams

A combined event is two or more results taken together: two dice thrown, a coin tossed and a spinner spun, two counters taken from a bag. The question is about the pair of results, not one of them on its own.
The rule you already have still works. If every outcome is equally likely, the probability is the number of outcomes you want divided by the total number of outcomes. The new difficulty is only in the counting: every outcome listed once, none missed and none counted twice.
A sample space is the list of every possible outcome. With a coin and a six-sided die, each of the 2 coin results can go with each of the 6 die results, so there are $2 \times 6 = 12$ outcomes: H1, H2, …, H6, T1, …, T6.
A list is fine for a small case. For two dice it gets messy, so you use a sample space diagram: a grid with the results of one event along the top, the results of the other down the side, and one cell for every pair.
Each cell is one equally likely outcome. Two fair dice give $6 \times 6 = 36$ cells. To find the probability of a total of 8, count the cells that show 8. There are 5, so the probability is $\frac{5}{36}$ .
graph diagram — not drawn yet
A 6 by 6 grid. Column headings 1 to 6 across the top, labelled 'First die'; row headings 1 to 6 down the left, labelled 'Second die'. Each cell holds the sum of its row and column headings, so the top row reads 2 3 4 5 6 7 and the bottom row reads 7 8 9 10 11 12. The five cells containing 8 are shaded.
36 equally likely cells. Five of them show a total of 8, so P(total = 8) = 5/36.
The trap here is to treat the totals $2, 3, \dots, 12$ as 11 equally likely outcomes and say the probability of 8 is $\frac{1}{11}$ . The totals are not equally likely. A total of 2 comes only from (1, 1), one cell. A total of 7 comes from six cells. The cells are equally likely; the totals are not.
The cells do not have to hold totals. Put in them whatever the question is about: the product, the difference, the larger number, or just the pair itself. Read the question first, then fill the grid.
When the two events have different numbers of outcomes, the grid is a rectangle. A spinner with 4 equal sections and a die give a $4 \times 6$ grid with 24 cells, and every probability from it starts with 24 as its denominator, before you simplify.

Part 2 of 6

Venn diagrams for probability

A Venn diagram sorts a group into regions. For two sets there are four regions: $A$ only, the overlap, $B$ only, and the space outside both circles. Every member of the group is in exactly one region.
A region can hold a count (how many people) or a probability. When it holds counts, a probability is the count in the region you want divided by the total for the whole rectangle. If 7 out of 40 students are in the overlap, the probability that a student chosen at random is in both sets is $\frac{7}{40}$ .
Read the notation as regions. $P (A \cap B)$ is the overlap only. $P (A \cup B)$ is everything inside either circle: $A$ only, the overlap and $B$ only. $P (A^{'})$ is everything outside circle $A$ , and that includes the region outside both circles.
graph diagram — not drawn yet
Three small two-set Venn diagrams side by side, each a rectangle labelled U with overlapping circles A (left) and B (right). Left, titled 'P(A ∩ B)': only the overlap shaded. Centre, titled 'P(A ∪ B)': both circles fully shaded, the outside unshaded. Right, titled 'P(A′)': the part of B outside A and the space outside both circles shaded, all of circle A unshaded.
∩ is the overlap only, ∪ is all of both circles, and A′ includes the space outside both circles.
To fill a diagram from totals, start in the overlap and work outwards. If 22 students play football and 7 of them also play tennis, the 'football only' region is $22 - 7 = 15$ , not 22. The 22 is the whole football circle, which already includes the 7 in the overlap.
Fill the outside region last: the total minus the three regions inside the circles. Then add all four regions. They must come to the total for the group, or to $1$ if the regions hold probabilities.
Sometimes you are given the number in neither set instead of the overlap. Then work the other way. If 30 people are in the group and 4 are in neither, $30 - 4 = 26$ are inside the circles. If the circle totals are 18 and 15, adding them gives $33$ , which is 7 more than 26, because the overlap was counted twice. So the overlap is $18 + 15 - 26 = 7$ . Now fill the rest from the overlap as usual.
The method is the same with probabilities. If $P (A) = 0.5$ , $P (B) = 0.4$ and $P (A \cap B) = 0.15$ , then $A$ only is $0.5 - 0.15 = 0.35$ and $B$ only is $0.4 - 0.15 = 0.25$ .
$P ((A \cup B)^{'}) = 1 - (0.35 + 0.15 + 0.25) = 0.25$

Part 3 of 6

Mutually exclusive events: when you may add

Two events are mutually exclusive if they cannot both happen at the same time. On one throw of a die, 'a 2' and 'a 5' are mutually exclusive, because the die cannot show both.
For mutually exclusive events, the probability that one or the other happens is the sum of their probabilities.
$P (A or B) = P (A) + P (B)$
Adding works because the two lists of outcomes have nothing in common, so the number of outcomes in ' $A$ or $B$ ' is just the two counts added. $P (2 or 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$ .
On a Venn diagram, mutually exclusive events are two circles that do not overlap, because $P (A \cap B) = 0$ .
The rule fails when the events can happen together. On one throw, let $A$ be 'an even number' and $B$ be 'a number greater than 3'. Then $P (A) = \frac{3}{6}$ and $P (B) = \frac{3}{6}$ . Adding gives $\frac{6}{6} = 1$ , which would mean $A$ or $B$ is certain. It is not: a 1 or a 3 is neither.
What went wrong is that 4 and 6 are in both lists, so adding counted them twice. List the outcomes instead: ' $A$ or $B$ ' is ${2, 4, 5, 6}$ , so the probability is $\frac{4}{6} = \frac{2}{3}$ . When events overlap, count from a list, a grid or a Venn diagram. Do not add.
One use of adding you already rely on: an event and its complement are mutually exclusive, and one of them must happen. So $P (A) + P (A^{'}) = 1$ , which is where $P (A^{'}) = 1 - P (A)$ comes from.

Part 4 of 6

Independent events: when you may multiply

Two events are independent if the result of one does not change the probability of the other. A coin landing heads does not change what a die will show, so 'heads' and 'a 6' are independent.
For independent events, the probability that both happen is the product of their probabilities.
$P (A and B) = P (A) \times P (B)$
Multiplying works because you need a fraction of a fraction. Half of all the coin-and-die outcomes start with heads, and one sixth of those have a 6. So the fraction with both is $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ . That matches the list of 12 outcomes, where only H6 is heads and a 6.
Check it against the two-dice grid. $P (a 6 on both dice) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$ , and the grid has exactly one cell, (6, 6), out of 36.
An 'and' answer is never bigger than either probability on its own, because both things happening is no easier than one of them happening. If your 'and' answer is bigger than one of the separate probabilities, you have added instead of multiplied.
The rule as written needs independence. When the first result changes the probabilities for the second, as when counters are taken from a bag and kept out, you still multiply, but by the changed probability for the second stage. The section on replacement shows how.

Part 5 of 6

Tree diagrams

A tree diagram shows a combined event one stage at a time, with one set of branches per stage. The syllabus describes the layout: the outcomes are written at the ends of the branches and the probabilities beside the branches.
The branches from one point cover every possible result of that stage, so their probabilities must add up to $1$ . If the probability that a bus is late is $0.2$ , the 'not late' branch is $1 - 0.2 = 0.8$ . Check each set of branches before you calculate anything.
graph diagram — not drawn yet
A two-stage tree diagram drawn left to right, with the headings 'Monday' and 'Tuesday' above the two stages. From a start point, a branch labelled 0.2 ends at the outcome 'Late' and a branch labelled 0.8 ends at 'Not late'. From each of these, a branch labelled 0.2 ends at 'Late' and a branch labelled 0.8 ends at 'Not late'. To the right of the four ends: 0.2 × 0.2 = 0.04, 0.2 × 0.8 = 0.16, 0.8 × 0.2 = 0.16, 0.8 × 0.8 = 0.64.
Outcomes at the ends, probabilities beside the branches. Multiply along each path; the four ends add to 1.
Each path from the start to an end is one combined outcome. Multiply along a path to get its probability. This is the 'and' rule: late on Monday and late on Tuesday is $0.2 \times 0.2 = 0.04$ .
Add the end results when more than one path gives what you want. Different paths are mutually exclusive, because you can only finish at one end, so adding them is allowed. 'Late on exactly one day' is two paths, late then not late, and not late then late.
$0.2 \times 0.8 + 0.8 \times 0.2 = 0.16 + 0.16 = 0.32$
Check that the four ends add up to $1$ , because one of the paths must happen: $0.04 + 0.16 + 0.16 + 0.64 = 1$ . If they do not, a branch is wrong.
For at least one, use the complement. 'Late on at least one day' is every path except not late, not late. So it is $1 - 0.8 \times 0.8 = 1 - 0.64 = 0.36$ . One subtraction replaces adding three paths.
A tree can have three stages. Three coins give $2 \times 2 \times 2 = 8$ ends, and each path multiplies three probabilities: $P (HHH) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ .

Part 6 of 6

With and without replacement

When you take an item from a bag, note it and put it back, the bag is the same for the second pick. That is with replacement. The two picks are independent, so the second-stage branches are the same as the first.
When you keep it out, the bag has changed. That is without replacement, and the second pick depends on the first. Someone who takes a sweet and eats it, or a team picked from a group of people, is without replacement, even if the question does not use the word.
Take a bag of 5 red and 3 blue counters, 8 in total. The first pick is red with probability $\frac{5}{8}$ and blue with probability $\frac{3}{8}$ . If the first counter is red and stays out, the bag holds 4 red and 3 blue, 7 in total.
So after a red, the second-stage branches are $\frac{4}{7}$ red and $\frac{3}{7}$ blue. After a blue, the bag holds 5 red and 2 blue, so the branches are $\frac{5}{7}$ red and $\frac{2}{7}$ blue.
graph diagram — not drawn yet
A two-stage tree diagram drawn left to right, with the headings 'First counter' and 'Second counter'. From the start, a branch labelled 5/8 ends at 'Red' and a branch labelled 3/8 ends at 'Blue'. From the first 'Red', branches labelled 4/7 and 3/7 end at 'Red' and 'Blue'. From the first 'Blue', branches labelled 5/7 and 2/7 end at 'Red' and 'Blue'. To the right of the four ends: RR 20/56, RB 15/56, BR 15/56, BB 6/56.
Without replacement every second-stage denominator is 7, and only the colour already taken loses one.
Two things change on the second stage. The denominator drops by 1 on every branch, because there is one item fewer. The numerator drops by 1 only for the item that was taken. Each pair of second-stage branches still adds to $1$ , for example $\frac{4}{7} + \frac{3}{7} = 1$ .
Then multiply along and add between paths exactly as before.
$P (both red) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$
Without a calculator, leave every end over the same denominator (here 56) until the last line. Adding $\frac{15}{56} + \frac{15}{56}$ is easy; adding $\frac{15}{56}$ to a fraction you have already simplified is not.
Give a probability as a fraction, a decimal or a percentage, and do not mix a fraction and a decimal in one answer.

Worked examples

Example 1. A fair spinner has four equal sections numbered 1, 2, 3 and 4. A fair six-sided die is numbered 1 to 6. The spinner is spun and the die is thrown, and the two numbers are added to give a score. (a) Draw a sample space diagram for the score. (b) Find the probability that the score is 7. (c) Find the probability that the score is 8 or more.

Show the working

There are 4 spinner results and 6 die results, so the grid has $4 \times 6 = 24$ equally likely cells. Put the spinner down the side (4 rows) and the die along the top (6 columns).
(a) Fill each cell with the row number plus the column number. Row 1 reads 2 3 4 5 6 7. Each row below is one more than the row above: row 2 reads 3 4 5 6 7 8, row 3 reads 4 5 6 7 8 9, row 4 reads 5 6 7 8 9 10.
graph diagram — not drawn yet
A 4 by 6 grid. Column headings 1 to 6 across the top, labelled 'Die'; row headings 1 to 4 down the left, labelled 'Spinner'. Each cell holds the sum of its headings: row 1 reads 2 3 4 5 6 7, row 2 reads 3 4 5 6 7 8, row 3 reads 4 5 6 7 8 9, row 4 reads 5 6 7 8 9 10. The four cells containing 7 are circled and the six cells containing 8, 9 or 10 are shaded.
24 cells. Four 7s, one in each row, and six cells of 8 or more.
(b) Count the 7s. There is one in each row: (1, 6), (2, 5), (3, 4) and (4, 3). That is 4 cells out of 24.
$P (score = 7) = \frac{4}{24} = \frac{1}{6}$
(c) '8 or more' means 8, 9 or 10. Row 1 has none, row 2 has one (8), row 3 has two (8, 9) and row 4 has three (8, 9, 10). That is $0 + 1 + 2 + 3 = 6$ cells.
$P (score ⩾ 8) = \frac{6}{24} = \frac{1}{4}$
Check the denominator is 24 both times. A denominator of 9, the number of different scores from 2 to 10, would be the 'scores are the outcomes' trap.

Answer: (a) The 4 by 6 grid above. (b) $\frac{1}{6}$ (c) $\frac{1}{4}$

Example 2. In a class of 40 students, 22 play football, 17 play tennis and 7 play both. $F$ is the set of students who play football and $T$ is the set who play tennis. A student is chosen at random. Find (a) $P (F \cap T)$ , (b) $P (F \cup T)$ , (c) the probability that the student plays exactly one of the two sports, (d) $P (F^{'})$ .

Show the working

Start in the overlap: 7 students play both, so 7 goes where the circles cross.
Football only is $22 - 7 = 15$ . Tennis only is $17 - 7 = 10$ .
Outside both circles is $40 - (15 + 7 + 10) = 40 - 32 = 8$ . Check: $15 + 7 + 10 + 8 = 40$ .
graph diagram — not drawn yet
A rectangle labelled U, with 40 written in its top right corner, containing two overlapping circles labelled F (left) and T (right). The number 15 is in the part of F outside T, 7 in the overlap, 10 in the part of T outside F, and 8 inside the rectangle outside both circles.
Overlap first, then each 'only' region, then the outside. The four numbers add to 40.
(a) $F \cap T$ is the overlap, 7 students.
$P (F \cap T) = \frac{7}{40}$
(b) $F \cup T$ is all three regions inside the circles: $15 + 7 + 10 = 32$ .
$P (F \cup T) = \frac{32}{40} = \frac{4}{5}$
Notice that $22 + 17 = 39$ , not 32. Adding the two circle totals counts the 7 in the overlap twice. These events are not mutually exclusive, so the adding rule does not apply.
(c) Exactly one sport means football only or tennis only, not the overlap: $15 + 10 = 25$ , so the probability is $\frac{25}{40} = \frac{5}{8}$ .
(d) $F^{'}$ is everything outside the football circle: tennis only and the outside region, $10 + 8 = 18$ . So $P (F^{'}) = \frac{18}{40} = \frac{9}{20}$ . Check with the complement: $1 - \frac{22}{40} = \frac{18}{40}$ .

Answer: (a) $\frac{7}{40}$ (b) $\frac{4}{5}$ (c) $\frac{5}{8}$ (d) $\frac{9}{20}$

Example 3. The probability that Rafael's bus is late on any school day is $0.2$ , independently of every other day. Draw a tree diagram for Monday and Tuesday. Then calculate the probability that the bus is late (a) on both days, (b) on exactly one of the two days, (c) on at least one of the two days.

Show the working

First stage, Monday: two branches, 'late' with $0.2$ beside it and 'not late' with $1 - 0.2 = 0.8$ .
Second stage, Tuesday: from each Monday end, draw the same pair of branches, $0.2$ and $0.8$ . They are the same because the days are independent.
There are four ends: late-late, late-not late, not late-late and not late-not late. Multiplying along each gives $0.04$ , $0.16$ , $0.16$ and $0.64$ , which add to $1$ .
(a) Late on both days is the single path late-late.
$0.2 \times 0.2 = 0.04$
(b) Exactly one late day is two paths, late-not late and not late-late. Multiply along each, then add.
$0.2 \times 0.8 + 0.8 \times 0.2 = 0.16 + 0.16 = 0.32$
(c) At least one late day is every path except not late-not late, so use the complement.
$1 - 0.8 \times 0.8 = 1 - 0.64 = 0.36$
Check (c) a second way by adding the three paths: $0.04 + 0.16 + 0.16 = 0.36$ . The two routes agree.

Answer: (a) $0.04$ (b) $0.32$ (c) $0.36$

Example 4. A bag contains 5 red counters and 3 blue counters. Two counters are taken at random, one after the other, without replacement. Calculate the probability that (a) both counters are the same colour, (b) at least one counter is blue.

Show the working

Draw the tree. First pick: red $\frac{5}{8}$ , blue $\frac{3}{8}$ .
After a red, 4 red and 3 blue are left, so the branches are $\frac{4}{7}$ and $\frac{3}{7}$ . After a blue, 5 red and 2 blue are left, so the branches are $\frac{5}{7}$ and $\frac{2}{7}$ .
Multiply along each path, keeping the denominator 56.
$RR = \frac{20}{56} RB = \frac{15}{56} BR = \frac{15}{56} BB = \frac{6}{56}$
Check: $20 + 15 + 15 + 6 = 56$ , so the four ends add to $1$ .
(a) Same colour is red-red or blue-blue. Add those two ends.
$\frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}$
(b) At least one blue is every end except red-red.
$1 - \frac{20}{56} = \frac{36}{56} = \frac{9}{14}$
Had you treated this as with replacement, red-red would have been $\frac{5}{8} \times \frac{5}{8}$ , which is a different and wrong value. The words 'without replacement' are the signal to change the second-stage fractions.

Answer: (a) $\frac{13}{28}$ (b) $\frac{9}{14}$

Where people go wrong

Mistake: Adding the probabilities along a path of a tree diagram.

Instead: Multiply along a path. Add only between the end results of different paths.

A path is one result and then another, which is the 'and' rule. Adding along a path can give an answer above $1$ , which no probability can be.

Mistake: Using the same fractions on the second stage when the items are not replaced.

Instead: Without replacement, every second-stage denominator is one less, and the numerator is one less for the item already taken.

The bag has changed. With 5 red and 3 blue, taking out a red leaves 4 red out of 7, not 5 out of 8.

Mistake: Working out only one order for 'one of each', for example red then blue.

Instead: Include both paths, red then blue and blue then red, and add them.

'One of each' does not say which comes first. Both paths end with one of each colour, so both belong in the answer.

Mistake: Finding $P (A \cup B)$ as $P (A) + P (B)$ when $A$ and $B$ can happen together.

Instead: Add only for mutually exclusive events. Otherwise add the three regions inside the circles of a Venn diagram, or count from a list or grid.

Outcomes in both $A$ and $B$ are counted once in $P (A)$ and again in $P (B)$ .

Mistake: Writing a set's total in its 'only' region of a Venn diagram.

Instead: Fill the overlap first, then subtract it from each set's total.

A set's total is the whole circle, overlap included. Writing 22 in 'football only' as well as 7 in the overlap counts those 7 students twice.

Mistake: Treating the totals of two dice, 2 to 12, as 11 equally likely outcomes.

Instead: Use the 36 cells of the sample space diagram as the equally likely outcomes.

A total of 7 comes from six cells and a total of 2 from only one, so the totals do not have equal probabilities.

Check yourself

Two fair coins are tossed. Find the probability of (a) two heads, (b) at least one head.
Answer
(a) $\frac{1}{4}$ : the outcomes are HH, HT, TH and TT, and only HH is two heads. (b) $1 - P (TT) = 1 - \frac{1}{4} = \frac{3}{4}$ .
Two fair six-sided dice are thrown. Find the probability that both dice show the same number.
Answer
The diagonal of the grid from (1, 1) to (6, 6) has 6 cells, so the probability is $\frac{6}{36} = \frac{1}{6}$ .
$A$ and $B$ are mutually exclusive, with $P (A) = 0.3$ and $P (B) = 0.45$ . Find $P (A or B)$ and the probability that neither happens.
Answer
$P (A or B) = 0.3 + 0.45 = 0.75$ . Neither: $1 - 0.75 = 0.25$ .
$A$ and $B$ are independent, with $P (A) = 0.6$ and $P (B) = 0.3$ . Find $P (A and B)$ and the probability that neither happens.
Answer
$P (A and B) = 0.6 \times 0.3 = 0.18$ . Neither: $P (A^{'}) \times P (B^{'}) = 0.4 \times 0.7 = 0.28$ .
A box holds 4 green counters and 6 yellow counters. Two are taken at random without replacement. Find the probability that both are green.
Answer
$\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$ .
$P (A) = 0.5$ , $P (B) = 0.4$ and $P (A \cap B) = 0.15$ . Find $P (A \cup B)$ and $P ((A \cup B)^{'})$ .
Answer
$A$ only is $0.35$ and $B$ only is $0.25$ , so $P (A \cup B) = 0.35 + 0.15 + 0.25 = 0.75$ and $P ((A \cup B)^{'}) = 1 - 0.75 = 0.25$ .
Why must the probabilities on the branches from one point of a tree diagram add up to $1$ ?
Answer
The branches from one point list every possible result of that stage, and exactly one of them happens.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A fair spinner has three equal sections numbered 1, 2 and 3. A fair six-sided die is numbered 1 to 6. The spinner is spun and the die is thrown. The score is the number on the spinner multiplied by the number on the die.

(a) Draw a sample space diagram showing every possible score. 2 marks No calculator

Mark scheme

B2 3 by 6 grid with all 18 products correct: 1 2 3 4 5 6 / 2 4 6 8 10 12 / 3 6 9 12 15 18 (rows or columns either way round); or B1 for at least 15 correct products

(b) Find the probability that the score is 6. 1 mark No calculator

Mark scheme

B1 3/18 oe, e.g. 1/6

(c) Find the probability that the score is greater than 10. 2 marks No calculator

Mark scheme

M1 4 cells identified (12, 12, 15, 18), or a fraction with numerator their count from their diagram and denominator 18
A1 4/18 oe, e.g. 2/9

Question 2. A box contains 6 milk chocolates and 4 dark chocolates. Kofi takes a chocolate at random and eats it. He then takes a second chocolate at random and eats it.

(a) Draw a tree diagram to show the two picks. Write the outcomes at the ends of the branches and the probabilities beside the branches. 3 marks No calculator

Mark scheme

B1 first pick: milk 6/10, dark 4/10 oe
B1 second pick after milk: milk 5/9, dark 4/9 oe
B1 second pick after dark: milk 6/9, dark 3/9 oe

(b) Calculate the probability that both chocolates are dark. 2 marks No calculator

Mark scheme

M1 4/10 × 3/9; FT their tree
A1 12/90 oe, e.g. 2/15

(c) Calculate the probability that Kofi eats one milk chocolate and one dark chocolate. 3 marks No calculator

Mark scheme

M1 6/10 × 4/9 or 4/10 × 6/9 seen; FT their tree
M1 adds the products for both orders, e.g. 24/90 + 24/90; or M2 for the complement method 1 − 6/10 × 5/9 − 4/10 × 3/9 oe
A1 48/90 oe, e.g. 8/15

(d) Explain why the probabilities on the second set of branches are different from those on the first. 1 mark No calculator

Mark scheme

B1 the first chocolate is eaten (not replaced), so only 9 are left and there is one fewer of the type taken oe

Question 3. In a group of 50 people, 28 own a bicycle and 19 own a car. 8 people own neither a bicycle nor a car. $B$ is the set of people who own a bicycle and $C$ is the set of people who own a car.

(a) Show that 5 people own both a bicycle and a car. 2 marks No calculator

Mark scheme

M1 50 − 8 = 42 own at least one oe, or an equation such as (28 − x) + x + (19 − x) + 8 = 50
A1 28 + 19 − 42 = 5, or x = 5 from correct working with every step shown

(b) A person is chosen at random from the group. Find $P (B \cap C^{'})$ . 1 mark No calculator

Mark scheme

B1 23/50 oe

(c) Find $P (B \cup C)$ . 1 mark No calculator

Mark scheme

B1 42/50 oe, e.g. 21/25

(d) Two people are chosen at random from the group, without replacement. Calculate the probability that both of them own a car. 2 marks

Mark scheme

M1 19/50 × 18/49 oe
A1 342/2450 oe, e.g. 171/1225, or 0.140 awrt (0.13959…)

How this is examined

Drawing a two-stage tree diagram, with or without replacement: A bag, box or two-day context. You Draw or Complete the tree, with the outcomes at the ends of the branches and the probabilities beside them, before any probability is asked for. Each pair of branches from one point adds to $1$ . Without replacement, every second-stage denominator is one less and the numerator is one less for the item taken (for example $\frac{6}{10}$ then $\frac{5}{9}$ ). Every branch carries a probability and every end carries an outcome.
Choosing which paths of the tree answer the question: A Calculate part after the tree asking for 'both', 'exactly one', 'one of each' or 'at least one'. A product written along each path you need (method), the products for both orders added when the order is not fixed (method), then one final fraction or decimal (accuracy). For 'at least one', writing $1 - P (none)$ with the numbers in shows the method.
Filling a Venn diagram from totals, then turning regions into probabilities: Totals for two groups plus the number in neither or in both. You Find or Show that the overlap has a value, then Find probabilities written as $P (A \cap B)$ , $P (A \cup B)$ or $P (A \cap B^{'})$ . Overlap first, each 'only' region as the set total minus the overlap, the outside region as the group total minus the three inside regions. Each probability is a region count over the total for the whole rectangle. For a Show that, the subtraction or equation that produces the given number, written out in full.
Counting cells in a sample space diagram: Two spinners, two dice, or a spinner and a die combined by a sum, product or difference. You Draw or Complete the grid, then Find probabilities from it. Every cell filled correctly. Each probability as (number of cells that fit) over (total number of cells), where the total is rows times columns, not the number of different values in the grid.
Deciding whether to add or multiply: A short part giving $P (A)$ and $P (B)$ and saying the events are mutually exclusive or independent, or an Explain asking why a rule does or does not apply. $P (A or B) = P (A) + P (B)$ used only for mutually exclusive events (they cannot both happen), and $P (A and B) = P (A) \times P (B)$ only for independent events (one does not change the probability of the other). An Explain states that reason for the events in the question.

Words to know

Combined event: Two or more results taken together, such as the scores on two dice or two counters taken from a bag.
Sample space: The list of every possible outcome.
Sample space diagram: A grid with one event's results along the top and the other's down the side, with one cell for each equally likely pair.
Mutually exclusive: Describes events that cannot both happen at the same time. For these, $P (A or B) = P (A) + P (B)$ .
Independent: Describes events where the result of one does not change the probability of the other. For these, $P (A and B) = P (A) \times P (B)$ .
Tree diagram: A diagram with one set of branches per stage, the outcomes at the ends of the branches and the probabilities beside them.
With replacement: Each item is put back before the next pick, so every pick has the same probabilities.
Without replacement: Each item is kept out, so the next pick has one item fewer and its probabilities change.
$P (A \cap B)$: The probability that A and B both happen: the overlap of the two circles on a Venn diagram.
$P (A \cup B)$: The probability that A or B or both happen: all of the regions inside either circle.

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