Combined Events and Tree Diagrams

AND — multiply

For independent events: $P(A \text{ and } B) = P(A) \times P(B)$

OR — add

For mutually exclusive events: $P(A \text{ or } B) = P(A) + P(B)$

Independent means one event does not affect the other. Two coin tosses are independent; drawing two cards without replacement is not.

Worked Example 1 — Both rules together

$P(\text{rain}) = 0.3$ on Saturday and $0.4$ on Sunday, independently. Find (a) $P(\text{rain both days})$, (b) $P(\text{rain on exactly one day})$.

①(a) AND, so multiply: $0.3 \times 0.4 = 0.12$

②(b) Two ways: rain Sat only, or rain Sun only.

③Sat only: $0.3 \times 0.6 = 0.18$

④Sun only: $0.7 \times 0.4 = 0.28$

⑤These are mutually exclusive, so add: $0.18 + 0.28 = 0.46$

A tree diagram lays out a two-stage (or three-stage) experiment so that no outcome is missed.

The four end probabilities add to $0.12 + 0.18 + 0.28 + 0.42 = 1$

Draw the first set of branches for the first event, writing the probability on each.
From the end of each branch, draw the second set.
Check that the probabilities on each pair of branches add to $1$.
Multiply along each path to find the probability of that combined outcome.
Add the end probabilities of all the paths that satisfy the event.
Check the four end probabilities total $1$.

The two tree-diagram rules

Multiply along the branches · Add between the branches

If an object is replaced before the next pick, the bag is back to its original state. The probabilities on the second set of branches are the same as the first.

Worked Example 2 — With replacement

A bag has $4$ red and $6$ blue counters. One is taken, its colour noted, and it is replaced. A second is then taken. Find (a) $P(\text{both red})$, (b) $P(\text{one of each colour})$.

①$P(\text{red}) = \dfrac{4}{10} = 0.4$ and $P(\text{blue}) = 0.6$, both times.

②(a) $P(\text{RR}) = 0.4 \times 0.4 = 0.16$

③(b) $P(\text{RB}) = 0.4 \times 0.6 = 0.24$

④$P(\text{BR}) = 0.6 \times 0.4 = 0.24$

⑤$P(\text{one of each}) = 0.24 + 0.24 = 0.48$

There are two ways to get one of each. Forgetting the second path is the single most common error in tree diagram questions.

The key change

Both the numerator and the denominator change for the second pick

Taking an object out and keeping it means there is one fewer item in the bag, and one fewer of whatever colour was removed. The events are dependent.

Worked Example 3 — Without replacement

A bag has $4$ red and $6$ blue counters. Two are taken without replacement. Find (a) $P(\text{both red})$, (b) $P(\text{both the same colour})$, (c) $P(\text{different colours})$.

①First pick: $P(\text{red}) = \dfrac{4}{10}$, $P(\text{blue}) = \dfrac{6}{10}$.

②After a red: $3$ red and $6$ blue remain, out of $9$.

③After a blue: $4$ red and $5$ blue remain, out of $9$.

④(a) $P(\text{RR}) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$

⑤$P(\text{BB}) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}$

⑥(b) $P(\text{same}) = \dfrac{12}{90} + \dfrac{30}{90} = \dfrac{42}{90} = \dfrac{7}{15}$

⑦(c) $P(\text{different}) = 1 - \dfrac{7}{15} = \dfrac{8}{15}$

Keep the denominator as $90$ until the very end. Cancelling each fraction separately makes the branches much harder to add.

Worked Example 4 — A three-colour bag

A bag has $5$ green, $3$ yellow and $2$ white beads. Two are taken without replacement. Find $P(\text{both green})$ and $P(\text{no green})$.

①Total $= 10$ beads.

②$P(\text{both green}) = \dfrac{5}{10} \times \dfrac{4}{9} = \dfrac{20}{90} = \dfrac{2}{9}$

③"No green" means both from the $5$ non-green beads.

④$P(\text{no green}) = \dfrac{5}{10} \times \dfrac{4}{9} = \dfrac{20}{90} = \dfrac{2}{9}$

(A coincidence here: there happen to be $5$ green and $5$ non-green.)

The shortcut

$P(\text{at least one}) = 1 - P(\text{none})$

Worked Example 5 — At least one

A bag has $7$ white and $3$ black balls. Three are taken without replacement. Find $P(\text{at least one black})$.

①The opposite of "at least one black" is "no black at all" — three whites.

②$P(\text{WWW}) = \dfrac{7}{10} \times \dfrac{6}{9} \times \dfrac{5}{8}$

③$= \dfrac{210}{720} = \dfrac{7}{24}$

④$P(\text{at least one black}) = 1 - \dfrac{7}{24} = \dfrac{17}{24}$

Listing all the cases with one, two or three blacks would take far longer.

Worked Example 6 — Finding an unknown number

A bag contains $n$ red and $5$ blue counters. Two are taken without replacement. The probability that both are blue is $\dfrac{2}{9}$. Find $n$.

①Total counters $= n + 5$.

②$P(\text{BB}) = \dfrac{5}{n+5} \times \dfrac{4}{n+4} = \dfrac{20}{(n+5)(n+4)}$

③Set equal to $\dfrac{2}{9}$: $\;\dfrac{20}{(n+5)(n+4)} = \dfrac{2}{9}$

④Cross-multiply: $2(n+5)(n+4) = 180$, so $(n+5)(n+4) = 90$.

⑤$n^2 + 9n + 20 = 90 \Rightarrow n^2 + 9n - 70 = 0$

⑥Factorise: $(n + 14)(n - 5) = 0$, so $n = 5$ or $n = -14$.

⑦A count cannot be negative, so $n = \mathbf{5}$.

Check: $10$ counters altogether; $\dfrac{5}{10} \times \dfrac{4}{9} = \dfrac{20}{90} = \dfrac{2}{9}$ ✓

Always discard the negative root and state why — a number of counters must be a positive whole number.

AND

Multiply — for independent events.

OR

Add — for mutually exclusive events.

Tree rule

Multiply along the branches, add between them.

Branch check

Each pair of branches must add to $1$.

End check

All the final probabilities must add to $1$.

With replacement

Second-stage probabilities are unchanged.

Without replacement

Both top and bottom of the fraction change.

Two ways

"One of each" always has two paths — do not forget the second.

At least one

Use $1 - P(\text{none})$.

Keep denominators

Do not cancel until you have added the branches.

Question 1

$P(A) = 0.4$ and $P(B) = 0.5$, and $A$ and $B$ are independent. Find $P(A \text{ and } B)$.

▶ Show solution

$P(A \text{ and } B) = 0.4 \times 0.5 = 0.2$

Question 2

A fair coin is tossed and a fair die rolled. Find $P(\text{a head and a } 6)$.

▶ Show solution

Independent events, so multiply:

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

Question 3

A bag has $3$ red and $7$ green counters. One is taken and replaced, then another is taken. Find (a) $P(\text{both red})$, (b) $P(\text{one of each})$.

▶ Show solution

$P(\text{red}) = 0.3$, $P(\text{green}) = 0.7$ each time.

(a) $0.3 \times 0.3 = 0.09$

(b) RG: $0.3 \times 0.7 = 0.21$; GR: $0.7 \times 0.3 = 0.21$

Total $= 0.42$

Question 4

A bag has $3$ red and $7$ green counters. Two are taken without replacement. Find $P(\text{both red})$.

▶ Show solution

$P(\text{RR}) = \dfrac{3}{10} \times \dfrac{2}{9} = \dfrac{6}{90} = \dfrac{1}{15}$

Question 5

Using the bag from Question 4, find $P(\text{both green})$ and $P(\text{different colours})$.

▶ Show solution

$P(\text{GG}) = \dfrac{7}{10} \times \dfrac{6}{9} = \dfrac{42}{90} = \dfrac{7}{15}$

$P(\text{same}) = \dfrac{6}{90} + \dfrac{42}{90} = \dfrac{48}{90} = \dfrac{8}{15}$

$P(\text{different}) = 1 - \dfrac{8}{15} = \dfrac{7}{15}$

Question 6

The probability that Ali passes his driving test first time is $0.65$. If he fails, the probability he passes second time is $0.8$. Find $P(\text{he passes within two attempts})$.

▶ Show solution

Two paths lead to passing: pass first time, or fail then pass.

Pass first: $0.65$

Fail then pass: $0.35 \times 0.8 = 0.28$

$P(\text{passes within two}) = 0.65 + 0.28 = 0.93$

Question 7

A fair coin is tossed three times. Find $P(\text{at least one head})$.

▶ Show solution

$P(\text{no heads}) = P(\text{TTT}) = \left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$

$P(\text{at least one head}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$

Question 8

A box has $8$ chocolates: $5$ milk and $3$ dark. Two are eaten at random. Find $P(\text{at least one dark})$.

▶ Show solution

$P(\text{no dark}) = P(\text{both milk}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$

$P(\text{at least one dark}) = 1 - \dfrac{5}{14} = \dfrac{9}{14}$

Question 9

The probability that a bus is late is $0.2$ each day, independently. Over three days, find (a) $P(\text{late all three days})$, (b) $P(\text{late exactly once})$.

▶ Show solution

(a) $0.2^3 = 0.008$

(b) Late on exactly one day — there are three ways (day 1, day 2 or day 3).

Each has probability $0.2 \times 0.8 \times 0.8 = 0.128$

Total $= 3 \times 0.128 = 0.384$

Question 10

A bag contains $6$ blue and $n$ yellow counters. Two counters are taken at random without replacement. The probability that both are blue is $\dfrac{1}{3}$.

(a) Form an equation in $n$. (b) Solve it to find $n$. (c) Find $P(\text{both yellow})$.

▶ Show solution

(a) Total counters $= 6 + n$.

$P(\text{BB}) = \dfrac{6}{6+n} \times \dfrac{5}{5+n} = \dfrac{30}{(6+n)(5+n)}$

Setting this equal to $\dfrac{1}{3}$:

$$\frac{30}{(6+n)(5+n)} = \frac{1}{3}$$

(b) Cross-multiply: $(6+n)(5+n) = 90$

$n^2 + 11n + 30 = 90$

$n^2 + 11n - 60 = 0$

$(n + 15)(n - 4) = 0$, so $n = 4$ or $n = -15$.

A count cannot be negative, so $n = \mathbf{4}$.

Check: $10$ counters; $\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}$ ✓

(c) $P(\text{both yellow}) = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$