Set Notation & Conditional Probability (Edexcel A-Level Mathematics): Revision Notes

Example: Shade $A \cup B$ Step 1: Shade $A$.

Step 2: Shade $B$.

Example: Shade $A' \cap B'$ In other words:

Shade the overlapping region of $A'$ and $B'$.

Example Question

Consider that

• $P(A) = 0.25$

• $P(B) = 0.4$

• $P(C) = 0.45$

• $P(A \cap B \cap C) = 0.1$ Given that:

• A and B are independent

• B and C are independent

• $A \cap B' \cap C = \emptyset$ (where B' denotes the complement of B)

Questions:

a) Draw a Venn diagram to illustrate the probabilities.

b) Find:

• $P(A' \cap (B' \cup C))$
• $P((A \cup B) \cap C)$ c) State, with reasons, whether events $A'$ and $C$ are independent.

Worked Solutions

a) Draw a Venn diagram to illustrate the probabilities.

Step 1: The Venn diagram is drawn with labelled probabilities in each region.

b) Find:

• $P(A' \cap (B' \cup C))$

Step 1: Understand the Event

We need to find the probability of the event $A′∩(B′∪C)$

This represents the regions where the complement of $A$ (i.e., everything outside $A$) overlaps with the union of $B'$ (the complement of $B'$) and $C$ (inside or outside $C$).

Step 2: Identify the Regions

We break down $B′∪C$ into the areas covered by $B'$ (the complement of $B$) or $C$, then find where this intersects with $A'$.

The relevant regions are:

• $A′∩B′$ (outside $A$ and outside $B$)
• $A′∩B∩C$ (outside $A$, inside $B$, and inside $C$)
• $A′∩B'∩C$ (outside $A$, outside $B$, and inside $C$)

Step 3: Add the Probabilities

To find $P(A' \cap (B' \cup C))$, sum the probabilities of the regions:

• $P(A′∩B′) = 0.27$
• $P(A′∩B∩C) = 0.08$
• $P(A′∩B'∩C) = 0.18$

Step 4: Calculate the Total

Add the probabilities of these regions:

$0.27+0.08+0.18=0.53$

Thus, $P(A' \cap (B' \cup C)) = :success[0.53]$

b) Find:

• $P((A \cup B) \cap C)$

Step 1: Understand the Event

We need to find the probability of $P((A \cup B) \cap C)$

This represents the event where either $A$ or $B$ occurs (the union of $A$ and $B$) and where $C$ also occurs at the same time (the intersection with $C$).

Step 2: Identify the Regions

To compute this, we focus on the overlap between $C$ and the union of $A$ and $B$.

This includes two key regions:

• $A∩B∩C$ (where all three events occur)
• $A'∩B∩C$ (where $B$ and $C$ occur, but $A$ does not).

Step 3: Add the Probabilities

We now add the probabilities of these two regions:

• $P(A∩B∩C) = 0.1$
• $P(A'∩B∩C) = 0.08$

Step 4: Calculate the Total

Summing these probabilities gives:

c) State, with reasons, whether events $A'$ and $C$ are independent.

Step 1: Understand the Condition for Independence

Two events are independent if the probability of their intersection equals the product of their individual probabilities.

This means we need to check if $P(A' \cap C) = P(A') \times P(C)$

Step 2: Calculate $P(A')$ and $P(C)$

$P(A')$ is the complement of $P(A)$

Therefore,

We are given $P(C) = 0.45$

Step 3: Multiply $P(A')$ and $P(C)$

Now, multiply the probabilities of $A'$ and $C$:

Step 4: Calculate $P(A' \cap C)$

The probability of $A' \cap C$ is found by summing the relevant regions:

Step 5: Compare the Results

Since $0.35 \neq 0.3375$

Comparison shows that $P(A') \times P(C) \neq P(A' \cap C)$.

Therefore, $A'$ and $C$ are not independent.

Example Question

Part (a): A sample space diagram is drawn showing all possible outcomes of the sum of the numbers when two four-sided dice are thrown.

Part (b): Given that at least one die lands on a 3, the probability that the sum on the two dice is exactly $5$ is calculated.

Part (c): A modelling assumption used in the calculations is stated.

Worked Solutions

a) Sample Space Diagram:

The sample space shows all possible sums (ranging from $2$ to $8$) for each possible pair of rolls between the two dice.

b) Conditional Probability:

The diagram highlights the outcomes where at least one die shows a $3$, excluding all cases where no $3's$ are present.

There are 7 possible outcomes remaining, with 2 of them summing to 5.

The probability is therefore $\frac{2}{7}$.

c) Modelling Assumption:

The assumption made is that the outcome on each die is equally likely, meaning the dice are fair.

Example Question

• A school has $75$ students in Year $12$.
• Of these students:
• $25$ study only humanities subjects ($H$).
• $37$ study only science subjects ($S$).
• $11$ students study both science and humanities subjects.
• Students are categorized into those who study humanities ($H$), science ($S$), and those who do not study either ($DNSH$ - Do Not Study Humanities, $DNSS$ - Do Not Study Science).

Questions

a) Draw a two-way table to represent this information.

b) Find the following probabilities:

• $P(S' \cap H')$
• $P(S|H')$
• $P(H|S')$

Worked Solutions

a) Draw a two-way table to represent this information.

The table is constructed with rows representing whether students study science ($SS$) or do not study science ($DNSS$).

The columns represent whether students study humanities ($SH$) or do not study humanities ($DNSH$).

The entries in the table represent the number of students in each category:

• $11$ students study both science and humanities.
• $25$ study only humanities.
• $37$ study only science.
• The total number of students is $75$.

b) Find the following probabilities:

• $P(S' \cap H')$ This is the probability of students who do not study science and do not study humanities.

Using the table

b) Find the following probabilities:

• $P(S|H')$ This is the probability of students studying science given that they do not study humanities.

Using the table

b) Find the following probabilities:

• $P(H|S')$ This is the probability of students studying humanities given that they do not study science.

Using the table