Conditional Probability

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Notes

What is Conditional Probability?

A **conditional probability** is the probability of an event A occurring **given that** event B has already occurred.
It is written as P(A | B), meaning 'probability of A given B'.
The probability is calculated from a **restricted set** of outcomes (only those where B has happened), not from all possible outcomes.
For example, picking a multiple of 3 from even numbers {2,4,6,8} gives P(multiple of $3 | even) = \frac{1}{4}$ .

Calculating Conditional Probabilities from Tables, Venn Diagrams, and Tree Diagrams

Conditional probabilities are often found using **two-way tables**, **Venn diagrams**, or **tree diagrams**.
From a Venn diagram: $P(A | B) = (number$ in A∩B) / (number in B).
From a two-way table: restrict to the row/column of the given condition, then find the required fraction.
Tree diagrams naturally show conditional probabilities along branches (e.g., second draw probabilities change after first draw without replacement).

Combined Conditional Probabilities (Without Replacement)

When drawing two or more items **without replacement**, the probability of the second event depends on the outcome of the first.
Multiply probabilities along branches: P(A then $B) = P(A) \times P(B | A)$ .
For example, drawing two red marbles from a bag of 4 red and 2 yellow: $P(RR) = (\frac{4}{6}) \times (\frac{3}{5}) = \frac{12}{30} = \frac{2}{5}$ .
Always adjust the total number of items and the number of favourable items after each draw.

Using the 'Or' Rule for Multiple Cases

When there are multiple ways to achieve a result (e.g., different colour orders), calculate each case separately and **add** the probabilities.
For two items of different colours from a bag of 10 yellow, 6 blue, 4 green: $P(different) = P(YB) + P(BY) + P(YG) + P(GY) + P(BG) + P(GB)$ .
Alternatively, use $1 - P(same colour) = 1 - [P(YY)+P(BB)+P(GG)]$ .
Remember that order matters: YB is different from BY.

Working with Algebraic Probabilities

When the number of items is unknown, use algebra to set up equations.
Let the unknown number be a variable (e.g., x blue marbles, x+3 red marbles).
Write the probability of the required outcome in terms of x and set it equal to the given probability.
Solve the resulting equation (often a quadratic) to find the value of x.

Common Pitfalls and Tips

Do **not** simplify fractions too early; keep a common denominator for easy addition.
Read carefully: 'given that' signals a conditional probability.
For 'at least one' problems, it is often easier to use 1 − P(none).
In tree diagrams, label branches clearly and check that probabilities on branches from the same node sum to 1.

Venn Diagram for Conditional Probability

Tree Diagram for Without Replacement

Two-Way Table Example

Algebraic Probability Setup

Practice questions

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1.A bag contains 4 red marbles and 2 yellow marbles. Behnaz picks two marbles at random without replacement. What is the probability that both marbles are red?
Easy
- A $\frac{2}{5}$
- B $\frac{1}{3}$
- C $\frac{4}{15}$
- D $\frac{2}{15}$
2.A box contains 7 black pens and 8 orange pens. Two pens are chosen at random without replacement. What is the probability that at least one orange pen is chosen?
Easy
- A $\frac{104}{105}$
- B $\frac{1}{105}$
- C $\frac{8}{15}$
- D $\frac{7}{15}$
3.A bag contains 5 blue marbles and 2 green marbles. Bryn picks one marble at random without replacement. If this marble is not green, he picks another marble at random without replacement. He continues until he picks a green marble. What is the probability that he picks a green marble on his first, second or third attempt?
Medium
- A $\frac{5}{7}$
- B $\frac{20}{21}$
- C $\frac{1}{21}$
- D $\frac{2}{7}$
4.Bag A contains 3 black balls and 2 white balls. Bag B contains 1 black ball and 3 white balls. A ball is taken at random from each bag. What is the probability that the two balls have different colours?
Hard
- A $\frac{9}{20}$
- B $\frac{11}{20}$
- C $\frac{1}{2}$
- D $\frac{3}{10}$
5.A bag contains 5 green counters and 4 red counters. Esme takes three counters at random from the bag without replacement. What is the probability that the three counters are all the same colour?
Hard
- A $\frac{5}{42}$
- B $\frac{1}{21}$
- C $\frac{4}{42}$
- D $\frac{1}{14}$
6.A group of 200 people were asked which city they would like to visit next. The table shows the results: London 50, Paris 48, New York 56, Tokyo 46. Two people are chosen at random from the group of 200. What is the probability that one person would like to visit London next and the other person would like to visit New York next? Give your answer as a percentage.
Medium
- A14%
- B28%
- C7%
- D56%
7.A box contains 15 red pencils, 8 yellow pencils and 2 green pencils. Two pencils are picked at random without replacement. What is the probability that at least one pencil is red?
Medium
- A $\frac{15}{25}$
- B $\frac{195}{300}$
- C $\frac{105}{300}$
8.The speeds of 200 cars are measured. The table shows frequencies: $0<v\le 20:16, 20<v\le 40:34, 40<v\le 45:62, 45<v\le 50:58, 50<v\le 60:26, 60<v\le 80:4$ . Two of the 200 cars are chosen at random. What is the probability that they both have a speed greater than 50 km/h?
Medium
- A $\frac{30}{200} \times \frac{29}{199}$
- B $\frac{30}{200} \times \frac{30}{200}$
- C $\frac{30}{200} \times \frac{29}{200}$
- D $\frac{30}{200} \times \frac{30}{199}$

Summary

## Conditional Probability (IGCSE Extended) ### Key Idea A **conditional probability** is the probability of an event occurring *given* that another event has already happened. The sample space is reduced to only the outcomes of the given event. ### Calculating Conditional Probabilities - **Without replacement**: The number of outcomes decreases after each draw. - **Tree diagrams** help visualise successive events. - **Formula**: P(A given B) = number of outcomes in A that are also in B divided by total outcomes in B. ### Combined Conditional Probabilities When events happen one after another without replacement, multiply probabilities along branches of a tree diagram, adjusting the counts. #### Example: Drawing two marbles without replacement - P(blue first) = (number of blue) / (total) - P(blue second | blue first) = (blue left) / (total left) - P(both blue) = P(blue first) × P(blue second | blue first) ### Multiple Possibilities Use the **OR rule**: add probabilities of mutually exclusive outcomes. - For two events: P(different colours) = P(YB) + P(BY) + P(YG) + P(GY) + P(BG) + P(GB) - For three events: P(all same) = P(YYY) + P(BBB) + P(GGG) ### Algebraic Problems Set up an equation using the given probability, solve for the unknown quantity (e.g., number of marbles). ### Common Question Types - Drawing cards, marbles, or counters without replacement. - Selecting people from a group (e.g., from a table). - Using Venn diagrams or two-way tables with "given that" phrasing. ### Tips - Do **not** simplify fractions too early; keep common denominators for addition. - Remember that order matters: AB is different from BA. - "At least one" often means 1 − P(none).

Key terms

Conditional Probability: The probability of an event occurring given that another event has already occurred. The sample space is restricted to the outcomes of the given event.
Without Replacement: When an item is selected and not returned to the set, so the total number of items and the number of items of each type decrease for subsequent selections.
Tree Diagram: A diagram used to show the probabilities of successive events, with branches representing possible outcomes and their probabilities.
OR Rule: The probability of either of two mutually exclusive events occurring is the sum of their individual probabilities.

Slides

What is Conditional Probability?

Conditional probability is the probability of an event happening given that another event has already occurred.
It focuses on a smaller, restricted set of outcomes.
For example, picking a multiple of 3 from even numbers only: P(multiple of $3 | even) = \frac{1}{4}$ .

A bag has 5 red and 3 blue marbles. You pick a marble and do not replace it. What is the probability that the second marble is blue given the first was blue?

$\frac{3}{8}$
$\frac{2}{7}$
$\frac{2}{8}$
$\frac{3}{7}$

Combined Conditional Probabilities

When events happen one after another without replacement, the probability of the second event depends on the first.
Multiply probabilities along the branches of a tree diagram, adjusting the numbers after each draw.
Example: P(red then $red) = (\frac{26}{52}) \times (\frac{25}{51})$ .

Two cards are drawn without replacement from a standard deck. The probability that both are kings $is ____ \times ____$ .

Using Tree Diagrams

Tree diagrams help visualise conditional probabilities for multiple events.
Each branch shows the probability of that outcome given previous events.
Multiply along branches to find the probability of a sequence.

Which of these are true about conditional probability?

The probability of A given B is written as P(A|B).
In 'without replacement' problems, the denominator stays the same.
Tree diagrams can be used to solve conditional probability problems.

The 'Or' Rule for Multiple Cases

When there are multiple ways to get a result (e.g., different orders), add their probabilities.
For example, P(one red and one $blue) = P(RB) + P(BR)$ .
Do not simplify fractions too early; it makes addition easier.

A bag has 4 red and 2 yellow marbles. Two are taken without replacement. What is P(one red and one yellow)?

$\frac{8}{15}$
$\frac{4}{15}$
$\frac{2}{5}$
$\frac{1}{3}$

Working with Algebraic Probabilities

Sometimes the number of items is unknown, e.g., x blue and x+3 red marbles.
Write probabilities in terms of x, then set up an equation from given probability.
Solve for x, discarding invalid solutions (e.g., negative).

If there are x blue marbles and 2x+3 total, $P(first blue) = ____$ .

Example: Both Beads Different Colours

Method $1: P(different) = 1 - P(same)$ .
Method 2: List all pairs of different colours and add their probabilities.
Remember that order matters: YB is different from BY.

Which statements are correct for 'without replacement' problems?

The probability of the second event depends on the first.
P(A then $B) = P(A) \times P(B)$ always.
You can use a tree diagram to organise probabilities.

Three Events Without Replacement

Extend the same method to three draws: multiply three probabilities.
Adjust numerator and denominator after each draw.
Example: $P(YYY) = (\frac{10}{20}) \times (\frac{9}{19}) \times (\frac{8}{18})$ .

A bag has 5 green and 4 red counters. Three are taken without replacement. What is P(all same colour)?

$\frac{5}{42}$
$\frac{1}{21}$
$\frac{5}{21}$
$\frac{1}{42}$

Key Takeaways

Conditional probability: probability given something else has happened.
Without replacement: adjust numbers after each draw.
Use tree diagrams and the 'or' rule for multiple cases.
Algebra can help find unknown quantities.

A bag contains 3 black and 2 white balls. Two balls are taken without replacement. What is P(black then white)?

$\frac{3}{10}$
$\frac{6}{25}$
$\frac{3}{5}$
$\frac{2}{5}$

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