IB Applications and Interpretation SL · Probability · Section 3 of 4

Conditional probability and independence of events

3 worked examples · By Álvaro López

Key idea

Sometimes you know something before calculating a probability: "I already know that the student is in the second year", "I already know that the customer paid by card". That information changes the probability, and conditional probability measures how much. If the information changes nothing, the events are independent. In the IB exam it appears with contingency tables, Venn diagrams and given data such as $P (A)$ , $P (B)$ and $P (A \cup B)$ .

Conditional probability

Imagine that a library has 100 users and you want the probability that a randomly chosen one has a membership card. If you are first told that the user is a minor, you no longer choose from the 100: you choose only from the minors. The information reduces the sample space, and that is all that conditional probability does.

The formula makes sense: from the set $B$ , which is now your whole universe, only the cases that are also in $A$ count, that is, $A \cap B$ . Dividing by $P (B)$ brings the universe back to probability 1.

Note that $P (A ∣ B)$ and $P (B ∣ A)$ are different numbers: the denominator changes.

Product rule

If you rearrange the definition to isolate $P (A \cap B)$ , you obtain a very useful form for calculating the probability that both things occur.

Conditional probability with a contingency table

A contingency table (or two-way table) divides the individuals according to two characteristics at once. With it you do not need the formula: you simply keep only the row or column of the condition and divide.

With the formula you obtain the same thing: the number of individuals in $A \cap B$ divided by the total of the table is $P (A \cap B)$ , and the total of $B$ divided by the total of the table is $P (B)$ ; when you divide, the total of the table cancels out.

Independent events

Two events are independent when knowing that one has occurred does not change the probability of the other.

The most convenient check in the exam is the last one: calculate $P (A \cap B)$ in one way and $P (A) \cdot P (B)$ in another, and compare them. If they are equal, the events are independent; if not, they are dependent. Always write down both values.

A useful consequence: if $A$ and $B$ are independent, so are $A$ and $B^{'}$ , $A^{'}$ and $B$ , and $A^{'}$ and $B^{'}$ . So, for example, $P (A \cap B^{'}) = P (A) \cdot P (B^{'})$ .

Worked examples

Example 1 Basic

Season-pass holders on a train

120 people are travelling in a carriage. The table classifies them by age and by whether or not they hold a travel pass.

	With pass	Without pass	Total
Young people	28	12	40
Adults	32	48	80
Total	60	60	120

One person is chosen at random from the carriage.

a) Given that the person is young, calculate the probability that they hold a pass.

b) Given that they hold a pass, calculate the probability that they are young.

c) Given that they do not hold a pass, calculate the probability that they are an adult.

Solution

a) The condition is "young", so the universe is the 40 people in the first row. Of these, 28 hold a pass: $P (pass ∣ young) = \frac{28}{40} = \frac{7}{10}$

b) Now the condition is "with pass", that is, the first column, with 60 people. Of these, 28 are young: $P (young ∣ pass) = \frac{28}{60} = \frac{7}{15}$

c) The condition is "without pass": 60 people, of whom 48 are adults: $P (adult ∣ without pass) = \frac{48}{60} = \frac{4}{5}$

Notice that parts a) and b) give different results: the numerator is the same (28), but the denominator changes with the condition.

Example 2 Medium

Two events with given data

In a neighbourhood, a household is chosen at random. Let $A$ be the event "has a dog" and $B$ the event "has a cat". It is known that $P (A) = \frac{3}{8} P (B) = \frac{1}{2} P (A \cup B) = \frac{5}{8}$

a) Calculate the probability that the household has a dog and a cat.

b) Given that the household has a cat, calculate the probability that it has a dog.

c) Given that the household has a dog, calculate the probability that it has a cat.

d) Are $A$ and $B$ independent?

Solution

a) Use $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ and solve for it: $\frac{5}{8} = \frac{3}{8} + \frac{1}{2} - P (A \cap B) \Rightarrow P (A \cap B) = \frac{7}{8} - \frac{5}{8} = \frac{1}{4}$

b) Divide by the probability of the condition, which is $B$ : $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{1/4}{1/2} = \frac{1}{2}$

c) Now the condition is $A$ : $P (B ∣ A) = \frac{P ( A \cap B )}{P ( A )} = \frac{1/4}{3/8} = \frac{2}{3}$

d) Compare $P (A \cap B)$ with $P (A) \cdot P (B)$ : $P (A) \cdot P (B) = \frac{3}{8} \cdot \frac{1}{2} = \frac{3}{16} \neq = \frac{1}{4} = P (A \cap B)$ They do not coincide, so $A$ and $B$ are not independent. This is consistent with the previous parts: $P (A ∣ B) = \frac{1}{2} \neq = \frac{3}{8} = P (A)$ . Having a cat makes it more likely that the household has a dog.

Example 3 Hard

Independent, but not mutually exclusive

In a café, a customer chosen at random orders a dessert (event $A$ ) and orders a coffee (event $B$ ). It is known that $P (A) = \frac{3}{10}$ , that $P (A \cup B) = \frac{13}{20}$ and that $A$ and $B$ are independent.

a) Calculate $P (B)$ .

b) Calculate $P (A \cap B)$ .

c) Calculate the probability that a customer orders a coffee, given that they did not order a dessert.

d) Are $A$ and $B$ mutually exclusive?

Solution

a) Since they are independent, $P (A \cap B) = P (A) \cdot P (B)$ . Substitute this into the formula for the union, with $P (B) = b$ : $\frac{13}{20} = \frac{3}{10} + b - \frac{3}{10} b$ $\frac{13}{20} - \frac{6}{20} = \frac{7}{10} b \Rightarrow \frac{7}{20} = \frac{7}{10} b \Rightarrow b = \frac{1}{2}$

b) $P (A \cap B) = P (A) \cdot P (B) = \frac{3}{10} \cdot \frac{1}{2} = \frac{3}{20}$ .

c) If $A$ and $B$ are independent, so are $A^{'}$ and $B$ . Therefore, knowing that they did not order a dessert does not change the probability of the coffee: $P (B ∣ A^{'}) = P (B) = \frac{1}{2}$ This can be checked with the definition: $P (A^{'} \cap B) = P (B) - P (A \cap B) = \frac{1}{2} - \frac{3}{20} = \frac{7}{20}$ and $P (A^{'}) = \frac{7}{10}$ , so $\frac{7/20}{7/10} = \frac{1}{2}$ .

d) Since $P (A \cap B) = \frac{3}{20} \neq = 0$ , the events are not mutually exclusive: some customers order a dessert and a coffee at the same time.

In the exam: you earn marks for writing $P (A \cap B) = P (A) \cdot P (B)$ as a consequence of independence and substituting it into the formula for the union; without that step you cannot reach the value of $P (B)$ . In part d) it is not enough to say "they are independent, so they are not mutually exclusive": use the value of $P (A \cap B)$ to justify it.

Common mistakes

Summary

$P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )}$ : the probability of $A$ when the universe is reduced to $B$ .
Product rule: $P (A \cap B) = P (A ∣ B) \cdot P (B) = P (B ∣ A) \cdot P (A)$ .
In a contingency table, the denominator is the total of the row or column of the condition.
$A$ and $B$ are independent if $P (A ∣ B) = P (A)$ , or if $P (B ∣ A) = P (B)$ , or if $P (A \cap B) = P (A) \cdot P (B)$ . Otherwise they are dependent.
If $A$ and $B$ are independent, so are $A$ and $B^{'}$ , $A^{'}$ and $B$ , and $A^{'}$ and $B^{'}$ .
Mutually exclusive ( $P (A \cap B) = 0$ ) is not the same as independent: with positive probabilities, two events are never both.
In the exam, write $P (A \cap B)$ and $P (A) \cdot P (B)$ to justify independence.

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