SL 4.6 Solve combined and conditional probability problems | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Statistics and Probability

Choosing the right rule, organising outcomes, and interpreting information

Probability describes how likely an event is, on a scale from 0 to 1. In this lesson, an event is a set of outcomes, such as drawing a red card or selecting a student who takes mathematics. Combined probability asks about events linked by words such as “and” or “or”. Conditional probability asks for the chance of an event when some information is already known. Begin by identifying what is being asked and what outcomes remain possible; then select a rule that matches the situation.

What you will learn

Calculate probabilities of combined events using addition and multiplication rules.
Use a tree diagram or table to organise outcomes in a probability problem.
Calculate and interpret conditional probabilities.
Distinguish mutually exclusive events from independent events.

1. Prior knowledge: events and probability

For equally likely outcomes, probability is the number of outcomes that meet the condition divided by the total number of outcomes. For example, a fair six-sided die has six equally likely results. The probability of rolling an even number is three outcomes out of six. A probability cannot be less than 0 or greater than 1.

The complement of an event is the event that it does not happen. If

A

is an event, its complement is written A'. Since either

A

happens or it does not, the probabilities add to 1. This is useful when counting the outcomes in A' is easier than counting those in

A

.

Before calculating, define the events clearly. For instance, let

A

mean “the selected student studies biology” and

B

mean “the student studies chemistry”. The event

A\cap B

means both conditions hold;

A\cup B

means at least one holds. The symbols describe the wording, not a new calculation method by themselves.

P(A')=1-P(A)

Check whether outcomes are equally likely before using a simple favourable-over-total count.
Use a complement when “not” or “at least one” is easier to count indirectly.
Translate “and” into an intersection and “or” into a union.

2. Combined events: “or” and “and”

For “

A

or

B

”, first check whether the events can happen together. If they are mutually exclusive, they cannot occur at the same time, so add their probabilities. If they can overlap, adding alone counts the overlap twice; subtract it once. This gives a reliable rule for any two events.

For “

A

and

B

”, the multiplication rule uses a conditional probability: the chance that both happen is the chance of

A

multiplied by the chance of

B

after

A

has happened. If the events are independent, knowing that

A

happened does not change the probability of

B

, and the multiplication becomes simpler.

A tree diagram is a useful representation for a sequence of events. Each branch is labelled with a probability, and probabilities along one complete path are multiplied. If several paths produce the requested event, add the path probabilities. In sampling without replacement, the probabilities on later branches change because the contents of the group have changed.

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Mutually exclusive events have no shared outcomes.
Independent events do not change each other’s probabilities.
On a tree, multiply along a path and add across relevant paths.

3. Conditional probability and representations

A conditional probability is written

P(A\mid B)

and read as “the probability of

A

given

B

”. The event after the vertical bar is the condition: the sample space is restricted to outcomes in

B

. In a two-way table, this means use the row or column that satisfies

B

as the new total, rather than the original total.

For example, if a survey records whether students walk to school and whether they bring lunch, a question about bringing lunch given that a student walks uses only the walkers as its denominator. The table provides a numerical representation; a fraction expresses the calculation; and the wording explains the context. Always state which group forms the denominator.

A conditional probability also links to the multiplication rule. Rearranging the definition gives the probability that both

A

and

B

occur. This relationship is especially helpful for a tree diagram or for finding a missing probability in a table.

A calculator can check arithmetic, especially for several branches or a large table. Enter the counts or probabilities only after deciding the correct denominator and rule. A graphing calculator does not decide whether events overlap, whether sampling is with replacement, or what the condition means. Show the event structure and a fraction or product in working, then use the calculator to evaluate it. Give a requested decimal to the stated accuracy.

P(A\mid B)=\frac{P(A\cap B)}{P(B)}

In $P(A\mid B)$ , restrict attention to outcomes where $B$ occurs.
The denominator is $P(B)$ , provided $P(B)>0$ .
Check a calculator result against the context: probabilities must lie between 0 and 1.

4. A decision routine for problems

Read the question carefully and mark the key word: “and”, “or”, “given”, “at least one”, or “neither”. Define events using short descriptions. Note whether outcomes are equally likely and whether a selection is with or without replacement. These details determine whether branch probabilities stay the same.

Choose a representation that makes the sample space visible. A short sequence often suits a tree; two categories may suit a table; a small equally likely set may be counted directly. Write the denominator explicitly for a conditional probability. For an “or” question, identify any overlap before adding.

Finally, calculate, report the answer in context, and check its size. For an “and” event, the result cannot exceed either individual event probability. For a conditional probability, compare the answer with the group specified by the condition, not with the whole population.

Define events before substituting values.
Use the representation that makes the relevant outcomes and denominator clear.
Check whether the result is reasonable in context.

Worked example

An “or” event with overlap

In a group of 40 students, 22 study Spanish, 18 study French, and 8 study both. Find the probability that a randomly selected student studies Spanish or French.

Define the events
Let $S$ mean that the student studies Spanish and $F$ mean that the student studies French. The question asks for at least one of these events.
Account for the overlap
Adding the two subject counts includes the 8 students who study both twice. Subtract that shared group once, then divide by all 40 students.
$P(S\cup F)=\frac{22+18-8}{40}$
Evaluate
There are 32 students in the union, so the probability is eight tenths.
$P(S\cup F)=\frac{32}{40}=0.8$

Answer: The probability is

0.8

, or 80%.

Check: The union count, 32, is at least as large as either subject count and no greater than 40.

Worked example

A conditional probability from a table

A club has 30 members. Of these, 12 are in Year 11, and 8 of the Year 11 members play chess. Find the probability that a member plays chess given that the member is in Year 11.

Identify the restricted group
The condition is that the member is in Year 11. Therefore, consider the 12 Year 11 members, not all 30 club members.
Form the conditional fraction
Of the 12 members in the condition group, 8 play chess. Use 12 as the denominator.
$P(\text{chess}\mid\text{Year 11})=\frac{8}{12}$
Simplify and interpret
The fraction simplifies to two thirds, which is approximately 0.667. Rounded to three significant figures, this is 0.667.
$P(\text{chess}\mid\text{Year 11})=\frac{2}{3}\approx0.667$

Answer: The probability is

\frac{2}{3}

, approximately

0.667

to three significant figures.

Check: The conditional probability uses only the Year 11 group; dividing 8 by 30 would answer a different question.

Worked example

Two selections without replacement

A bag contains 5 green counters and 3 yellow counters. Two counters are drawn at random without replacement. Find the probability that the first is green and the second is yellow.

Set up the first branch
There are 8 counters at the first draw, and 5 are green. The probability of green first is five eighths.
$P(G_1)=\frac{5}{8}$
Update the second probability
Given a green first draw, 7 counters remain and all 3 yellow counters remain. The chance of yellow second is three sevenths.
$P(Y_2\mid G_1)=\frac{3}{7}$
Multiply along the path
The required sequence follows one tree path, so multiply the probability of green first by the conditional probability of yellow second.
$P(G_1\cap Y_2)=\frac{5}{8}\times\frac{3}{7}=\frac{15}{56}$

Answer: The probability is

\frac{15}{56}

, approximately

0.268

to three significant figures.

Check: The second denominator is 7 because the first counter is not replaced.

Common mistakes and how to avoid them

Adding

P(A)

and

P(B)

for every “or” question.

Correction: Check for overlap. If both events can happen together, subtract

P(A\cap B)

once.

Using the original total as the denominator in a conditional probability.

Correction: Use only the outcomes satisfying the condition after the vertical bar.

Treating successive draws without replacement as independent.

Correction: Update the number and type of remaining outcomes after each draw.

Adding probabilities along a tree path.

Correction: Multiply along one path; add the probabilities of separate paths that meet the requested condition.

Lesson summary

For an “or” event, add the probabilities and subtract any overlap.
For an “and” event, multiply a probability by the relevant conditional probability.
For a condition, restrict the sample space to the group named after “given”.
Use tree diagrams and tables to keep outcomes, changing probabilities, and denominators clear.

Check your understanding

Question 1

A fair die is rolled. What is the probability of rolling a number greater than 4 or an even number?

$\frac{2}{3}$
$\frac{1}{2}$
$\frac{5}{6}$
$\frac{1}{3}$

Show answer and explanation

\frac{2}{3}

The events are

\{5,6\}

and

\{2,4,6\}

. Their union is

\{2,4,5,6\}

, giving

\frac{4}{6}=\frac{2}{3}

.

Question 2

If

P(A\cap B)=0.12

and

P(B)=0.4

, what is

P(A\mid B)

?

$0.30$
$0.48$
$0.12$
$0.52$

Show answer and explanation

0.30

Divide the joint probability by the probability of the condition:

0.12\div0.4=0.30

.

Question 3

A fair coin is tossed twice. What is the probability of tails on both tosses?

$\frac{1}{4}$
$\frac{1}{2}$
$\frac{3}{4}$
$1$

Show answer and explanation

\frac{1}{4}

The tosses are independent, so multiply the two tail probabilities:

\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}

.

Key terms

Event: A specified set of outcomes of a probability situation.
Mutually exclusive: Events that cannot happen at the same time.
Independent: Events where knowing that one happened does not change the probability of the other.
Conditional probability: The probability of an event when it is known that another event has occurred.
Complement: The event consisting of all outcomes where a specified event does not occur.

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SL 4.6 · Solve combined and conditional probability problems

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Choosing the right rule, organising outcomes, and interpreting information

What you will learn

1. Prior knowledge: events and probability

2. Combined events: “or” and “and”

3. Conditional probability and representations

4. A decision routine for problems

An “or” event with overlap

A conditional probability from a table

Two selections without replacement

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review