Elementary Probability

Lessons

Senior 6 · Basic Mathematics

Elementary Probability

5 lessons

Lesson 3 of 5

Independent events, the multiplication rule and tree diagrams

25 min

"And" in probability

A school basketball player scores 70% of her free throws (an illustrative figure). She is fouled and takes two throws. What is the probability that she scores both? If the first throw makes no difference to the second, she scores the first in 70% of such situations, and in 70% of those she scores the second as well: $0.7 \times 0.7 = 0.49$ . Probabilities of events that do not influence each other are multiplied.

Definition

Two events $A$ and $B$ are independent if the occurrence of one does not change the probability of the other. Examples: the scores on two dice; tossing a coin and rolling a die; drawing two cards with replacement.
Formally, $A$ and $B$ are independent if and only if

P(A \cap B) = P(A) \times P(B)

A tree diagram shows a sequence of experiments: each branch is labelled with its probability, the branches from one point add up to 1, and the probability of a path is the product of the probabilities along it.

The multiplication rule for independent events

Why it works. For two experiments with equally likely outcomes, the product rule from Unit 2 counts the combined sample space: if the first has $n_{1}$ outcomes and the second $n_{2}$ , together they have $n_{1} n_{2}$ . If $A$ uses $a$ outcomes of the first and $B$ uses $b$ outcomes of the second, then $A \cap B$ contains $a b$ of the combined outcomes, so

P(A \cap B) = \frac{ab}{n_1 n_2} = \frac{a}{n_1} \times \frac{b}{n_2} = P(A) \times P(B)

The rule extends to any number of independent events: $P(A \cap B \cap C) = P(A)\,P(B)\,P(C)$ .

Tree diagrams

Here are the two free throws (S = scores, M = misses), assuming the throws are independent:

Multiply along a path to get the probability of that outcome. Add down the end column to combine outcomes that are mutually exclusive. For example, "exactly one score" is SM or MS:

P(\text{exactly one}) = 0.21 + 0.21 = 0.42

Check: the four end probabilities add to $0.49 + 0.21 + 0.21 + 0.09 = 1$ .

Worked example

Worked example 1: "at least one". Using the tree, find the probability that the player scores at least one of the two throws.

The only outcome with no score is MM, so

P(\text{at least one}) = 1 - P(MM) = 1 - 0.3^2 = 1 - 0.09 = 0.91

In general, if each of $n$ independent trials succeeds with probability $p$ , then $P(\text{at least one success}) = 1 - (1 - p)^n$ .

Worked example

Worked example 2: with replacement. A bag contains 5 red and 3 blue counters. A counter is drawn, its colour noted, and it is put back; then a second counter is drawn. Find the probability that both counters are the same colour.

Because the first counter is replaced, the draws are independent and the probabilities stay $\frac{5}{8}$ and $\frac{3}{8}$ on the second draw.

P(\text{same}) = P(RR) + P(BB) = \left(\frac{5}{8}\right)^2 + \left(\frac{3}{8}\right)^2 = \frac{25}{64} + \frac{9}{64} = \frac{34}{64} = \frac{17}{32}

Worked example

Worked example 3: testing for independence. $P (A) = 0.4$ , $P (B) = 0.5$ and $P(A \cap B) = 0.2$ . Are $A$ and $B$ independent? What if $P(A \cap B) = 0.3$ ?

$P(A) \times P(B) = 0.4 \times 0.5 = 0.2 = P(A \cap B)$ , so $A$ and $B$ are independent.

If $P(A \cap B) = 0.3 \ne 0.2$ , the events are dependent: knowing that one happened changes the chance of the other.

Worked example

Worked example 4: three machines. A small factory has three machines that break down independently, each with probability 0.1 on a given day. Find the probability that on a given day (a) all three work; (b) at least one breaks down.

(a) $P(\text{all work}) = 0.9^3 = 0.729$ .

(b) $P(\text{at least one breaks down}) = 1 - 0.729 = 0.271$ .

Warning

Independent is not the same as mutually exclusive. Mutually exclusive events cannot happen together, so $P(A \cap B) = 0$ ; if both have non-zero probability, they are certainly dependent, because one happening makes the other impossible. Independent events can, and usually do, happen together.

Tip

On a tree diagram: along the branches, multiply; down the outcomes, add. And check that every set of branches from one point adds up to 1.

Khan Academy: "Compound probability of independent events". Notice how each new independent event multiplies the probability again.

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Practice

A fair coin is tossed and a fair die is rolled. Find $P(\text{head and a 6})$ and $P(\text{tail and an even number})$ .
$A$ and $B$ are independent with $P (A) = 0.3$ and $P (B) = 0.6$ . Find $P(A \cap B)$ and $P(A \cup B)$ .
Three fair dice are rolled. Find the probability that all three show a six.
A fair die is rolled four times. Find the probability of at least one six, to 3 decimal places.
A bag has 4 green and 6 white beads. Two are drawn with replacement. Draw a tree diagram and find the probability of one of each colour.
A bus is late on any given day with probability 0.2, independently of other days. Find the probability that in a school week of 5 days it is never late, to 3 significant figures.
$P (A) = 0.5$ , $P (B) = 0.3$ , $P(A \cup B) = 0.65$ . Are $A$ and $B$ independent? Show your working.

Show answersHide answers

$\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ ; $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ .
$0.3 \times 0.6 = 0.18$ ; $0.3 + 0.6 - 0.18 = 0.72$ .
$\left(\frac{1}{6}\right)^3 = \frac{1}{216}$ .
$1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518$ .
$P(GW) + P(WG) = 0.4 \times 0.6 + 0.6 \times 0.4 = 0.48$ .
$0.8^5 = 0.32768 \approx 0.328$ .
$P(A \cap B) = 0.5 + 0.3 - 0.65 = 0.15$ and $P (A) P (B) = 0.15$ , so yes, they are independent.

Important

$A$ and $B$ are independent $\iff P(A \cap B) = P(A)\,P(B)$ . On a tree, multiply along a path and add the paths you want. "At least one" $= 1 - P(\text{none})$ .

Sources and credits

Text, tree diagram, worked examples and exercises: Written by Ganzaa. The free-throw and machine figures are illustrative.

Video: Compound probability of independent events, Khan Academy, embedded from YouTube.

1"And" in probability

2The multiplication rule for independent events

3Tree diagrams

"And" in probability

The multiplication rule for independent events

Tree diagrams