| Representation | Use it when… |
|---|---|
| Tree diagram | There are successive stages, especially with changing probabilities |
| Two-way table | Each item is classified by two attributes at once |
| Venn diagram | Categories overlap, and you need "both", "neither" or "only one" |
| Binomial distribution | The same trial is repeated many times with a fixed success probability |
A bag holds $5$ red and $4$ blue counters. Two are taken without replacement. Find the probability of one of each colour.
Two machines make components. A makes $60\%$ with a $4\%$ fault rate; B makes $40\%$ with a $9\%$ fault rate. A component is faulty. Find the probability it came from B.
A two-way table classifies every item by two attributes. Its great strength is that the row and column totals check each other.
Of $200$ students, $120$ study maths, $90$ study physics and $40$ study neither. Build a two-way table and find how many study both.
| Physics | No physics | Total | |
|---|---|---|---|
| Maths | $50$ | $70$ | $120$ |
| No maths | $40$ | $40$ | $80$ |
| Total | $90$ | $110$ | $200$ |
$A'$ — not in $A$ · $n(A)$ — the number of elements in $A$ · $\xi$ — the universal set
Using the diagram above, find (a) $n(M \cup P)$, (b) the number studying exactly one subject, (c) $P(M \mid P)$ for a randomly chosen student.
- Are there successive stages, with probabilities that change? $\to$ tree diagram.
- Is each item classified two ways, with totals given? $\to$ two-way table.
- Do the words "both", "neither" or "only" appear? $\to$ Venn diagram.
- Is the same trial repeated $n$ times with fixed probability $p$? $\to$ binomial distribution.
In a club of $50$, $30$ play tennis, $25$ play badminton and $10$ play both. Find the probability a random member plays neither.
Along a branch
Multiply.
Between branches
Add.
All paths
Must total $1$.
"One of each"
Two paths — remember both orders.
Without replacement
The denominator drops by one.
Tables
Fill the totals first, then work inwards.
Venn
Fill the overlap first.
Addition rule
$n(A \cup B) = n(A)+n(B)-n(A\cap B)$.
Conditional
Wanted path $\div$ all paths meeting the condition.
Many stages
Switch to the binomial distribution.
A coin is tossed twice. List all the outcomes and state the probability of exactly one head.
▶ Show solution
HH, HT, TH, TT — four equally likely outcomes.
Exactly one head: HT and TH, so $P = \dfrac{2}{4} = 0.5$.
A bag has $3$ red and $7$ green counters. Two are taken with replacement. Find $P(\text{both red})$.
▶ Show solution
With replacement, the probabilities do not change.
$P = \dfrac{3}{10} \times \dfrac{3}{10} = \dfrac{9}{100} = 0.09$
Repeat Question 2 without replacement.
▶ Show solution
After one red is removed, $2$ reds remain out of $9$.
$P = \dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{6}{90} = \dfrac{1}{15}$
In a class of $30$, $18$ have a bike, $14$ have a scooter and $6$ have both. How many have neither?
▶ Show solution
$n(B \cup S) = 18 + 14 - 6 = 26$
Neither: $30 - 26 = 4$
Of $150$ people, $80$ are female and $95$ own a car, of whom $50$ are female. Complete a two-way table.
▶ Show solution
| Car | No car | Total | |
|---|---|---|---|
| Female | $50$ | $30$ | $80$ |
| Male | $45$ | $25$ | $70$ |
| Total | $95$ | $55$ | $150$ |
Male total $= 150 - 80 = 70$; male car owners $= 95 - 50 = 45$; the rest follow.
A bag has $4$ white and $6$ black balls. Two are drawn without replacement. Find $P(\text{one of each colour})$.
▶ Show solution
WB: $\dfrac{4}{10} \times \dfrac{6}{9} = \dfrac{24}{90}$
BW: $\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}$
$P = \dfrac{48}{90} = \dfrac{8}{15}$
$P(A) = 0.4$, $P(B) = 0.5$, $P(A \cap B) = 0.2$. Find $P(A \cup B)$ and $P(A' \cap B')$.
▶ Show solution
$P(A \cup B) = 0.4 + 0.5 - 0.2 = 0.7$
$P(A' \cap B')$ means "neither", which is the complement of "at least one":
$= 1 - 0.7 = 0.3$
Two factories supply parts. X supplies $70\%$ with $3\%$ faulty; Y supplies $30\%$ with $8\%$ faulty. Find the overall fault rate.
▶ Show solution
$P(X \text{ and faulty}) = 0.7 \times 0.03 = 0.021$
$P(Y \text{ and faulty}) = 0.3 \times 0.08 = 0.024$
$P(\text{faulty}) = 0.045$, i.e. $4.5\%$.
Explain why a tree diagram would be a poor choice for finding the probability of exactly $5$ heads in $12$ coin tosses.
▶ Show solution
A tree diagram for $12$ tosses would have $2^{12} = 4096$ end branches — impossible to draw.
Worse, you would need to identify the ones with exactly $5$ heads, and there are $\binom{12}{5} = 792$ of them.
The binomial distribution does this in one line: $P(X=5) = \binom{12}{5}(0.5)^5(0.5)^7 = 792 \times (0.5)^{12} = \dfrac{792}{4096} = 0.193$ (3 s.f.).
The tree diagram is the right tool for two or three stages; beyond that, use the binomial.
A screening test for a condition affecting $3\%$ of a population correctly detects it $92\%$ of the time, and wrongly gives a positive result for $6\%$ of healthy people.
(a) Using a population of $10\,000$, build a two-way table. (b) Find $P(\text{positive test})$. (c) Find $P(\text{has the condition} \mid \text{positive test})$. (d) Comment on what this means in practice.
▶ Show solution
(a) With the condition: $3\%$ of $10\,000 = 300$; without: $9700$.
True positives: $92\%$ of $300 = 276$, leaving $24$ false negatives.
False positives: $6\%$ of $9700 = 582$, leaving $9118$ true negatives.
| Positive | Negative | Total | |
|---|---|---|---|
| Has condition | $276$ | $24$ | $300$ |
| No condition | $582$ | $9118$ | $9700$ |
| Total | $858$ | $9142$ | $10\,000$ |
(b) $P(\text{positive}) = \dfrac{858}{10\,000} = 0.0858$
(c) Restrict to the $858$ positives, of whom $276$ genuinely have it:
$P = \dfrac{276}{858} = 0.322$ (3 s.f.)
(d) Only about a third of people who test positive actually have the condition — two in three positives are false alarms.
This happens because the condition is rare: a $6\%$ error rate applied to $9700$ healthy people produces $582$ false positives, which swamps the $276$ true ones. The test is still valuable as a first screen, catching $92\%$ of real cases, but every positive result must be confirmed by a more specific second test before anyone is told they have the condition.