so $P(\text{event}) = \dfrac{\text{number of cells that work}}{\text{total number of cells}}$
Two fair coins are tossed. Draw a sample space diagram and find $P(\text{one head and one tail})$.
| H | T | |
|---|---|---|
| H | HH | HT |
| T | TH | TT |
The most important sample space at GCSE is the grid of totals when two dice are rolled. It is worth being able to reproduce it quickly.
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Total | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Ways | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $5$ | $4$ | $3$ | $2$ | $1$ |
Two fair dice are rolled and the scores added. Find (a) $P(\text{total} = 9)$, (b) $P(\text{total} \geq 10)$, (c) $P(\text{total is even})$.
Two dice are rolled. Find $P(\text{the two scores are the same})$ and $P(\text{the difference is } 1)$.
- Put the outcomes of the first experiment down the side.
- Put the outcomes of the second along the top.
- Fill each cell with the combined result (a total, a product, a pair of letters…).
- Count the total number of cells.
- Count the cells that satisfy the event, and divide.
A three-sided spinner numbered $1$, $2$, $3$ is spun and a fair four-sided die numbered $1$–$4$ is rolled. The two numbers are multiplied. Find $P(\text{product} \gt 6)$.
| × | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| 1 | 1 | 2 | 3 | 4 |
| 2 | 2 | 4 | 6 | 8 |
| 3 | 3 | 6 | 9 | 12 |
Spinner A has sections R, B, G. Spinner B has sections $1$, $2$. Both are spun. List the sample space and find $P(\text{a red with an even number})$.
Two dice are rolled. Find $P(\text{the total is a prime number})$.
A biased spinner has $P(\text{red}) = 0.6$ and $P(\text{blue}) = 0.4$. It is spun twice. Find $P(\text{two reds})$ and $P(\text{one of each})$.
Check: $P(\text{BB}) = 0.16$, and $0.36 + 0.24 + 0.24 + 0.16 = 1$ ✓
Sample space
The complete list of possible outcomes.
Grid method
First experiment down the side, second along the top.
Two dice
$36$ outcomes; totals from $2$ to $12$.
Most likely total
$7$, with $6$ ways out of $36$.
Order matters
HT and TH are two different outcomes.
Counting
$P = \dfrac{\text{cells that work}}{\text{total cells}}$.
Check the total
Number of cells $=$ outcomes of first $\times$ outcomes of second.
Biased?
Do not count cells — multiply the individual probabilities.
Read carefully
"Greater than" excludes the number; "at least" includes it.
Two fair coins are tossed. Find (a) $P(\text{two tails})$, (b) $P(\text{at least one head})$.
▶ Show solution
Sample space: HH, HT, TH, TT — four outcomes.
(a) One outcome: $\dfrac{1}{4}$
(b) Three outcomes (HH, HT, TH): $\dfrac{3}{4}$
Two fair dice are rolled and the scores added. Find (a) $P(\text{total} = 5)$, (b) $P(\text{total} = 12)$.
▶ Show solution
(a) $(1,4),(2,3),(3,2),(4,1)$ — four ways: $\dfrac{4}{36} = \dfrac{1}{9}$
(b) Only $(6,6)$: $\dfrac{1}{36}$
Two fair dice are rolled. Find $P(\text{total is less than } 6)$.
▶ Show solution
Totals $2$, $3$, $4$, $5$ give $1 + 2 + 3 + 4 = 10$ ways.
$P = \dfrac{10}{36} = \dfrac{5}{18}$
Two fair dice are rolled. Find $P(\text{both scores are even})$.
▶ Show solution
Each die has $3$ even faces, so the grid has $3 \times 3 = 9$ favourable cells.
$P = \dfrac{9}{36} = \dfrac{1}{4}$
A four-sided spinner ($1$–$4$) and a coin are used together. Draw the sample space and find $P(\text{a head with an odd number})$.
▶ Show solution
Outcomes: H1, H2, H3, H4, T1, T2, T3, T4 — eight in total.
Head with odd number: H1, H3 — two outcomes.
$P = \dfrac{2}{8} = \dfrac{1}{4}$
Two fair dice are rolled and the scores multiplied. Find $P(\text{product is } 12)$.
▶ Show solution
Pairs giving $12$: $(2,6), (3,4), (4,3), (6,2)$ — four ways.
$P = \dfrac{4}{36} = \dfrac{1}{9}$
Two dice are rolled. Find $P(\text{the difference between the scores is } 3)$.
▶ Show solution
Pairs: $(1,4),(4,1),(2,5),(5,2),(3,6),(6,3)$ — six ways.
$P = \dfrac{6}{36} = \dfrac{1}{6}$
A biased coin has $P(\text{heads}) = 0.7$. It is tossed twice. Find (a) $P(\text{two heads})$, (b) $P(\text{exactly one head})$.
▶ Show solution
The outcomes are not equally likely, so multiply.
(a) $0.7 \times 0.7 = 0.49$
(b) HT: $0.7 \times 0.3 = 0.21$; TH: $0.3 \times 0.7 = 0.21$
Total $= 0.42$
A spinner numbered $1$, $2$, $3$ is spun twice and the two numbers are added.
(a) Draw the sample space of totals. (b) Find $P(\text{total} = 4)$. (c) Find $P(\text{total is odd})$.
▶ Show solution
(a) The grid of totals:
| + | 1 | 2 | 3 |
|---|---|---|---|
| 1 | 2 | 3 | 4 |
| 2 | 3 | 4 | 5 |
| 3 | 4 | 5 | 6 |
Nine outcomes in total.
(b) Total $4$ appears three times: $P = \dfrac{3}{9} = \dfrac{1}{3}$
(c) Odd totals are $3$ (twice) and $5$ (twice) — four cells.
$P = \dfrac{4}{9}$
In a game, two fair dice are rolled. You win £$5$ if the total is $10$ or more, £$2$ if the total is exactly $7$, and nothing otherwise. It costs £$1$ to play.
(a) Find $P(\text{winning £5})$. (b) Find $P(\text{winning nothing})$. (c) Find the expected winnings per game. (d) Is the game fair?
▶ Show solution
(a) Totals $10$, $11$, $12$: $3 + 2 + 1 = 6$ ways.
$P = \dfrac{6}{36} = \dfrac{1}{6}$
(b) Total $7$ has $6$ ways, so winning something covers $6 + 6 = 12$ cells.
$P(\text{nothing}) = \dfrac{36 - 12}{36} = \dfrac{24}{36} = \dfrac{2}{3}$
(c) Expected winnings $= \left(5 \times \dfrac{6}{36}\right) + \left(2 \times \dfrac{6}{36}\right) + \left(0 \times \dfrac{24}{36}\right)$
$= \dfrac{30}{36} + \dfrac{12}{36} = \dfrac{42}{36} = £1.17$ (to the nearest penny)
(d) The game costs £$1$ but returns about £$1.17$ on average, so it is not fair — it favours the player by about $17$p per game. The organiser would lose money in the long run.