Independent Events and Tree Diagrams

When two things happen one after another — tossing a coin twice, drawing two counters with replacement, checking the weather on two days — we often want the probability that both happen. If the result of the first does not affect the second, the events are independent, and we use the AND (multiplication) rule. The clearest tool for organising these problems is the tree diagram, where you multiply along the branches and add between paths. This lesson covers independence, the multiplication rule, and how to build and read tree diagrams for events with replacement.

This is AO1 (applying the multiplication rule and constructing tree diagrams), AO2 (deciding whether events are independent), and AO3 (multi-step "at least one" problems combining rules). OCR command words include Work out, Calculate, Complete (the tree) and Show that.

---

Key Vocabulary

Term	Definition
Independent	Two events where the outcome of one does not affect the other.
With replacement	An item is put back before the next selection, so probabilities stay the same.
AND (multiplication) rule	For independent events, $P(A \text{ and } B) = P(A) \times P(B)$P(A and B)=P(A)×P(B).
Tree diagram	A branching diagram showing the outcomes and probabilities of successive events.
Branch	One line of a tree showing a single outcome and its probability.
Path	A route through the tree giving one combined outcome.

---

Independent Events and the AND Rule

Two events are independent if the outcome of the first has no effect on the second. Tossing a coin twice, rolling two dice, or drawing a counter, replacing it, and drawing again are all independent. For independent events,

$P(A \text{ and } B) = P(A) \times P(B)$P(A and B)=P(A)×P(B)

Why multiply? Think of it as a fraction of a fraction. If $A$A happens on a fraction $P(A)$P(A) of occasions, and then within those occasions $B$B happens a fraction $P(B)$P(B) of the time, the proportion of occasions on which both happen is $P(A)$P(A) of $P(B)$P(B) — and "of" means multiply. Tossing two heads is a half of a half, which is a quarter; this matches the sample space $\{HH, HT, TH, TT\}${HH,HT,TH,TT}, where $HH$HH is one outcome out of four.

The contrast between "and" and "or" is worth fixing firmly in your mind, because mixing them up is the commonest error in this whole topic. The word "and" (both events happening together) calls for multiplication of independent probabilities, and the result is smaller than either probability — requiring two things makes an outcome less likely. The word "or" (either event happening) calls for addition of mutually exclusive probabilities, and the result is larger. A quick self-check after any calculation: if you multiplied and got a bigger number, or added and got a smaller one, you have almost certainly used the wrong rule.

Worked Example 1

A fair coin is tossed twice. Work out the probability of getting two heads.

Solution: The tosses are independent.

$P(\text{H and H}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$P(H and H)=21​×21​=41​

Worked Example 2

A fair dice is rolled and a fair coin is tossed. Work out the probability of getting a $6$6 and a head.

Solution: $P(6 \text{ and head}) = \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}$P(6 and head)=61​×21​=121​

Worked Example 3

A bag contains $3$3 red and $2$2 blue counters. A counter is taken, its colour noted, and replaced. A second counter is then taken. Work out the probability that both are red.

Solution: Because the counter is replaced, the draws are independent and $P(\text{red}) = \dfrac{3}{5}$P(red)=53​ each time.

$P(\text{R and R}) = \dfrac{3}{5} \times \dfrac{3}{5} = \dfrac{9}{25}$P(R and R)=53​×53​=259​

Common error: Changing the second probability to $\dfrac{2}{4}$42​. That is for without replacement; here the counter is replaced, so it stays $\dfrac{3}{5}$53​.

---

Drawing a Tree Diagram

A tree diagram shows each stage of an experiment as a set of branches. The probability of each outcome is written on its branch, and the branches from any single point add up to $1$1. Below is a two-stage tree for tossing a fair coin twice (H = heads, T = tails).

graph LR
    S["Start"] -->|"1/2"| H1["H (1st)"]
    S -->|"1/2"| T1["T (1st)"]
    H1 -->|"1/2"| HH["H, H"]
    H1 -->|"1/2"| HT["H, T"]
    T1 -->|"1/2"| TH["T, H"]
    T1 -->|"1/2"| TT["T, T"]

The two rules for using a tree:

• Multiply along the branches of a single path to find the probability of that combined outcome (AND).
• Add the probabilities of separate paths to find the probability of "one path or another" (OR).

These two rules are just the AND and OR rules from before, now organised visually. Reading from left to right along a single path corresponds to "this happens and then that happens", which is why you multiply; jumping between different complete paths corresponds to "either this whole sequence or that one", which is why you add. Because the different end-points of a tree are mutually exclusive (you can only end up at one of them), adding their probabilities is always valid.

A tree diagram has three properties worth checking every time. First, the branches leaving any single point must sum to $1$1, since something must happen at each stage — this is a quick way to catch a mislabelled branch. Second, multiplying along all the paths and adding the results must give $1$1, because the end-points are exhaustive. Third, the structure makes "at least one" problems easy: the path with no successes is usually a single product, so $1 - P(\text{none})$1−P(none) is fast. Getting into the habit of these checks turns the tree from a drawing into a reliable calculating machine.

Worked Example 4

Using the coin tree above, work out the probability of getting exactly one head.

Solution: "Exactly one head" happens on the paths H,T and T,H.

$P(\text{H,T}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}, \qquad P(\text{T,H}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$P(H,T)=21​×21​=41​,P(T,H)=21​×21​=41​

$P(\text{exactly one head}) = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$P(exactly one head)=41​+41​=21​

Worked Example 5

The probability that it rains on any day is $0.3$0.3, independently of other days. Draw a tree for two days and work out the probability it rains on both days.

Solution: Each day has $P(\text{rain}) = 0.3$P(rain)=0.3 and $P(\text{no rain}) = 0.7$P(no rain)=0.7.

graph LR
    S["Start"] -->|"0.3"| R1["Rain (Day 1)"]
    S -->|"0.7"| N1["No rain (Day 1)"]
    R1 -->|"0.3"| RR["Rain, Rain"]
    R1 -->|"0.7"| RN["Rain, No rain"]
    N1 -->|"0.3"| NR["No rain, Rain"]
    N1 -->|"0.7"| NN["No rain, No rain"]

$P(\text{rain on both days}) = 0.3 \times 0.3 = 0.09$P(rain on both days)=0.3×0.3=0.09

---

Combining Branches: "At Least One"

Many questions ask for the probability of "at least one" success. The fastest method is usually the complement: $P(\text{at least one}) = 1 - P(\text{none})$P(at least one)=1−P(none).

The reason the complement is so useful here is that "at least one" hides a lot of separate cases. "At least one head in three tosses" covers exactly one head, exactly two heads, and three heads — three different situations, each needing several paths added up. But the opposite of "at least one" is simply "none", a single, easy outcome: no heads at all. So instead of adding many path probabilities, you compute the one probability of getting none and subtract it from $1$1. The more trials there are, the bigger the saving, which is why experienced students reach for the complement the moment they see the words "at least one".

You can always check the complement answer the long way if you have time: add up the probabilities of all the favourable paths and confirm you get the same value. The two methods must agree, because the favourable paths and the "none" path together make up the whole tree, whose probabilities sum to $1$1.

Worked Example 6

Using the rain tree above, work out the probability that it rains on at least one of the two days.

Solution: $P(\text{no rain on either day}) = 0.7 \times 0.7 = 0.49$P(no rain on either day)=0.7×0.7=0.49 $P(\text{at least one rainy day}) = 1 - 0.49 = 0.51$P(at least one rainy day)=1−0.49=0.51

Common error: Adding $0.3 + 0.3 = 0.6$0.3+0.3=0.6. "At least one" is not a single AND or a single OR — use the complement, or add the three favourable paths ($RR + RN + NR = 0.09 + 0.21 + 0.21 = 0.51$RR+RN+NR=0.09+0.21+0.21=0.51).

Worked Example 7

A spinner lands on red with probability $0.4$0.4. It is spun twice. Work out the probability of getting at least one red.

Solution: $P(\text{not red}) = 0.6$P(not red)=0.6 each spin.