Fair — all outcomes are equally likely. A fair die has $P = \tfrac{1}{6}$ for each face.
Biased — the outcomes are not equally likely. A weighted die might land on six far too often.
| Situation | Fair or biased? |
|---|---|
| An ordinary six-sided die | Assumed fair unless told otherwise |
| A drawing pin dropped on a table | Biased — point-up and point-down are not equally likely |
| A spinner with unequal sectors | Biased — bigger sectors are more likely |
| Names drawn from a hat, well mixed | Fair |
| A coin balanced on its edge and nudged | Could be biased — depends on the method |
If you know the probability of an event, you can predict roughly how many times it will happen in a large number of trials.
- Work out the probability of the event.
- Identify how many trials there will be.
- Multiply the two together.
- Round to a whole number if the context requires it (people, cars, etc.).
A fair die is rolled $180$ times. How many fours would you expect?
The probability that a bus is late is $0.18$. In a month there are $250$ bus journeys. How many would you expect to be late?
A spinner is spun $400$ times and is expected to land on blue $120$ times. Find $P(\text{blue})$.
The probability of winning a game is $0.04$. How many games must be played to expect $10$ wins?
A machine produces components. The probability a component is faulty is $0.03$. On Monday it makes $2000$ components and on Tuesday $3500$. How many faulty components would you expect in total?
A coin is tossed twice, and this pair of tosses is repeated $200$ times. How many times would you expect to get two heads?
In a game, the probability of winning £$5$ is $0.1$, and otherwise you win nothing. It costs £$1$ to play. If you play $300$ times, how much would you expect to win or lose overall?
You would expect to lose about £$150$. The game is not fair to the player.
- Find the probability of each possible prize.
- Multiply each prize by its probability.
- Add these together to get the expected winnings for one go.
- Compare with the cost of one go.
A game costs £$2$ to play. You spin a spinner: $P(\text{£}10) = 0.05$, $P(\text{£}3) = 0.2$, and otherwise you win nothing. Is the game fair?
The game is not fair — it favours the organiser by $90$p per play.
Check the probabilities: $0.05 + 0.2 + 0.75 = 1$ ✓
Random
Every item has the same chance of being chosen.
Fair
All outcomes equally likely, so $P = \dfrac{\text{favourable}}{\text{total}}$ applies.
Biased
Outcomes are not equally likely; you need experimental data.
Expected number
$P \times$ number of trials.
Finding $P$
Divide the expected number by the number of trials.
Finding $n$
Divide the expected number by the probability.
Not a guarantee
The expected number is a long-run average, not a certainty.
Fair game
Expected winnings equal the cost of a go.
A fair coin is tossed $340$ times. How many tails would you expect?
▶ Show solution
$P(\text{tails}) = 0.5$
Expected $= 0.5 \times 340 = 170$ tails
A fair die is rolled $420$ times. How many times would you expect a score greater than $4$?
▶ Show solution
More than $4$ means $5$ or $6$, so $P = \dfrac{2}{6} = \dfrac{1}{3}$.
Expected $= \dfrac{1}{3} \times 420 = 140$ times
The probability that a seed germinates is $0.85$. A gardener plants $360$ seeds. How many would you expect to germinate, and how many to fail?
▶ Show solution
Germinate: $0.85 \times 360 = 306$
Fail: $360 - 306 = 54$ (or $0.15 \times 360 = 54$)
A biased spinner lands on red with probability $0.28$. It is spun $150$ times. How many times would you expect it to land on red? Give your answer to the nearest whole number.
▶ Show solution
$0.28 \times 150 = 42$ times
A machine is expected to produce $84$ faulty items out of every $2400$. Find the probability that an item is faulty.
▶ Show solution
$P = \dfrac{84}{2400} = 0.035$
That is $3.5\%$.
The probability of a raffle ticket winning is $0.008$. How many tickets must be sold to expect $6$ winners?
▶ Show solution
$0.008 \times n = 6$
$n = \dfrac{6}{0.008} = 750$ tickets
Explain why dropping a drawing pin is not a fair experiment, and how you could estimate the probability that it lands point up.
▶ Show solution
The two outcomes (point up and point down) are not equally likely — the shape and weight distribution of the pin favour one of them. So you cannot say $P = \tfrac{1}{2}$.
To estimate the probability, drop the pin a large number of times (say $500$), record how many times it lands point up, and calculate the relative frequency $\dfrac{\text{point up}}{500}$.
A coin is tossed three times, and this is repeated $400$ times. How many times would you expect to get three heads?
▶ Show solution
$P(\text{HHH}) = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$
Expected $= \dfrac{1}{8} \times 400 = 50$ times
A game costs £$1$ to play. The probability of winning £$4$ is $0.15$ and the probability of winning £$1$ is $0.25$; otherwise you win nothing.
(a) Find the expected winnings per go. (b) Is the game fair? (c) If $600$ people play, what profit would the organiser expect?
▶ Show solution
(a) $P(\text{nothing}) = 1 - 0.15 - 0.25 = 0.6$
Expected winnings $= (4 \times 0.15) + (1 \times 0.25) + (0 \times 0.6)$
$= 0.60 + 0.25 = £0.85$
(b) The cost is £$1$ but the expected return is £$0.85$, so the game is not fair — the player loses $15$p per go on average.
(c) Profit $= 600 \times £0.15 = £90$
A four-sided spinner has sections numbered $1$, $2$, $3$ and $4$. The probabilities are shown, with $P(3) = P(4)$.
| Number | $1$ | $2$ | $3$ | $4$ |
|---|---|---|---|---|
| Probability | $0.36$ | $0.22$ | $x$ | $x$ |
(a) Find $x$. (b) The spinner is spun $250$ times. How many $3$s would you expect? (c) Is the spinner fair? Explain. (d) How many spins would be needed to expect $100$ scores of $1$?
▶ Show solution
(a) All probabilities add to $1$:
$0.36 + 0.22 + x + x = 1$
$0.58 + 2x = 1 \Rightarrow 2x = 0.42 \Rightarrow x = 0.21$
(b) $0.21 \times 250 = 52.5$, so about $\mathbf{53}$ threes (you cannot have half a spin, so round sensibly).
(c) No. A fair four-sided spinner would have $P = 0.25$ for every number. Here the probabilities are $0.36$, $0.22$, $0.21$, $0.21$ — not equal, so the spinner is biased towards $1$.
(d) $0.36 \times n = 100$
$n = \dfrac{100}{0.36} = 277.8$, so about $\mathbf{278}$ spins.