CIE AS/A Level Maths-5.3 Probability- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-5.3 Probability- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-5.3 Probability- Study Notes
Key Concepts:
- Evaluating Probabilities by Enumeration, Permutations & Combinations
- Use Of Addition And Multiplication Of Probabilities
- Exclusive and Independent events
- Conditional Probability:
Evaluating Probabilities by Enumeration, Permutations & Combinations
Evaluating Probabilities by Enumeration, Permutations & Combinations
Sample space (S): the set of all possible elementary outcomes. If all elementary outcomes are equally likely, then for an event \(E\),
\( \displaystyle P(E)=\dfrac{\text{number of favourable outcomes}}{\text{number of outcomes in }S} \).
Enumeration:
For small finite experiments (e.g. throwing dice, tossing coins) list the sample space explicitly and count favourable outcomes.
Using permutations:
Use when order matters. Number of ordered ways to choose \(r\) items from \(n\) is \(^{\,n}P_r=\dfrac{n!}{(n-r)!}\).
Using combinations:
Use when order does not matter. Number of unordered ways to choose \(r\) items from \(n\) is \(^{\,n}C_r=\dfrac{n!}{r!(n-r)!}\).
Choosing method:
Decide whether order matters. If not, use combinations. If the sample space is equiprobable, probabilities can be computed as ratios of counts found via permutations/combinations.
Common checks:
- Sum of probabilities of disjoint outcomes = 1.
- For “without replacement” problems use combinations/permutations on the reduced sample space; for “with replacement” use counting with full sample space each draw.
Example
Two fair six-sided dice are thrown. Find:
- (a) the probability the total score is 7;
- (b) the probability both dice show 6.
▶️ Answer/Explanation
Sample space: ordered pairs \((i,j)\) with \(i,j\in\{1,\dots,6\}\). Total outcomes \(|S|=6\times6=36\) (all equiprobable).
(a) Total = 7 occurs for pairs: \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) → 6 favourable outcomes.
\( \displaystyle P(\text{sum }7)=\dfrac{6}{36}=\dfrac{1}{6}.\)
(b) Both dice show 6 is the single outcome \((6,6)\) → 1 favourable.
\( \displaystyle P(\text{both 6})=\dfrac{1}{36}.\)
\(\boxed{P(\text{sum }7)=\dfrac{1}{6},\quad P(\text{both 6})=\dfrac{1}{36}}\)
Example
A bag contains 3 red and 2 blue balls (5 balls total). Two balls are drawn at random without replacement. Find:
- (a) the probability both are red;
- (b) the probability exactly one is red.
▶️ Answer/Explanation
Method: Order does not matter for these events → use combinations. Total unordered pairs: \(^{5}C_{2}=\dfrac{5!}{2!3!}=10.\)
(a) Choose 2 red from 3: \(^{3}C_{2}=3\) favourable.
\( \displaystyle P(\text{both red})=\dfrac{^{3}C_{2}}{^{5}C_{2}}=\dfrac{3}{10}.\)
(b) Exactly one red = choose 1 red from 3 and 1 blue from 2: \(^{3}C_{1}\times{}^{2}C_{1}=3\times2=6\) favourable.
\( \displaystyle P(\text{exactly one red})=\dfrac{6}{10}=\dfrac{3}{5}.\)
(Check: \(3/10+3/5+(1/10)=1\) if include both-blue case \(^{2}C_{2}=1\).)
\(\boxed{P(\text{both red})=\dfrac{3}{10},\quad P(\text{exactly one red})=\dfrac{3}{5}}\)
Example
Same bag (3 red, 2 blue). Two balls are drawn with order recorded (first, then second). Find the probability the first is red and the second is blue.
▶️ Answer/Explanation
Method 1 (conditional probabilities):
\(P(\text{first red})=\dfrac{3}{5}.\) Given first is red, remaining balls: 2 red, 2 blue → \(P(\text{second blue}\mid \text{first red})=\dfrac{2}{4}=\dfrac{1}{2}.\)
So \( \displaystyle P(\text{R then B})=\dfrac{3}{5}\times\dfrac{1}{2}=\dfrac{3}{10}.\)
Method 2 (ordered counting): Number of ordered favourable sequences = \(3\times2=6\) (choose which red first (3 options) and which blue second (2 options)). Total ordered outcomes = \(5\times4=20\). So \(6/20=3/10\).
\(\boxed{P(\text{first red, second blue})=\dfrac{3}{10}}\)
Example
A standard 52-card deck. If 5 cards are dealt, what is the probability of getting exactly 2 aces?
▶️ Answer/Explanation
Total 5-card hands: \(^{52}C_{5}\).
Number of ways to have exactly 2 aces = choose 2 aces from 4 and 3 non-aces from 48: \(^{4}C_{2}\times{}^{48}C_{3}.\)
So
\( \displaystyle P(\text{exactly 2 aces})=\dfrac{^{4}C_{2}\times{}^{48}C_{3}}{^{52}C_{5}}.\)
(Leave in combinatorial form or evaluate numerically if needed.)
\(\boxed{P=\dfrac{^{4}C_{2}\,^{48}C_{3}}{^{52}C_{5}}\text{ (exact form)}}\)
Use Of Addition And Multiplication Of Probabilities
Use Of Addition And Multiplication Of Probabilities
Probabilities are used to describe the likelihood of events.
Addition Rule (for mutually exclusive events): If two events A and B cannot happen at the same time, then:
$\rm{P(A \cup B) = P(A) + P(B)}$
Multiplication Rule (for independent events): If the outcome of one event does not affect the other, then:
$\rm{P(A \cap B) = P(A) × P(B)}$
For events that are not mutually exclusive, we must be careful to avoid double counting:
$\rm{P(A \cup B) = P(A) + P(B) – P(A and B)}$
In simple A Level problems, you may be asked to apply these rules in real-life contexts such as dice, cards, or random selections.
Example:
A card is drawn at random from a standard deck of 52 cards. Find the probability that the card is either a Heart or a King.
▶️ Answer/Explanation
Total cards = 52
P(Heart) = 13/52, P(King) = 4/52
But note: one King is a Heart → overlap = 1/52
So, P(Heart or King) = P(Heart) + P(King) – P(Heart and King)
= (13/52) + (4/52) – (1/52) = 16/52 = \(\dfrac{4}{13}\)
Final Answer: \(\boxed{\dfrac{4}{13}}\)
Example:
Two fair dice are rolled. Find the probability that both dice show a 6.
▶️ Answer/Explanation
For each die: P(6) = 1/6
Dice are independent → use multiplication rule:
P(6 on both dice) = (1/6) × (1/6) = 1/36
Final Answer: \(\boxed{\dfrac{1}{36}}\)
Example:
A bag contains 5 red balls and 3 blue balls. Two balls are drawn without replacement. Find the probability that both are red.
▶️ Answer/Explanation
Total balls = 8
P(First red) = 5/8
After one red is taken: 4 red left out of 7
P(Second red) = 4/7
So, P(both red) = (5/8) × (4/7) = 20/56 = 5/14
Final Answer: \(\boxed{\dfrac{5}{14}}\)
Exclusive and Independent events
Exclusive and Independent events
Mutually Exclusive Events: Two events are mutually exclusive if they cannot occur at the same time.
If A and B are mutually exclusive: $\rm{P(A \cap B) = 0}$
Example: When rolling a die, A = “getting a 2”, B = “getting a 5”. Both cannot happen together.
Independent Events: Two events are independent if the outcome of one does not affect the other.
If A and B are independent: $\rm{P(A \cap B) = P(A) × P(B)}$
Example: Tossing a coin and rolling a die.
Key difference:
- Exclusive events → cannot happen together (intersection = 0).
- Independent events → can happen together, but probabilities multiply.
Example
A single die is rolled. Let event A = “getting an even number” and event B = “getting a 3”. Are A and B mutually exclusive?
▶️ Answer/Explanation
A = {2, 4, 6}, P(A) = 3/6
B = {3}, P(B) = 1/6
A ∩ B = {} (no overlap)
So, P(A ∩ B) = 0 → A and B are mutually exclusive.
Example
A coin is tossed and a die is rolled. Let event A = “getting a Head” and event B = “rolling a 6”. Are A and B independent?
▶️ Answer/Explanation
P(A) = 1/2, P(B) = 1/6
P(A ∩ B) = probability of (Head and 6) = (1/2) × (1/6) = 1/12
Check independence: P(A) × P(B) = (1/2) × (1/6) = 1/12
Since P(A ∩ B) = P(A) × P(B), A and B are independent.
Example
A bag contains 3 red and 2 blue balls. One ball is drawn without replacement. Let event A = “first ball is red”, event B = “second ball is red”. Are A and B independent?
▶️ Answer/Explanation
P(A) = 3/5
If first is red, then probability of second red = 2/4 = 1/2
So, P(A ∩ B) = (3/5) × (1/2) = 3/10
But P(A) × P(B) = (3/5) × (3/5) = 9/25 = 0.36
Since P(A ∩ B) ≠ P(A) × P(B), A and B are NOT independent.
Conditional Probability:
Conditional Probability:
Conditional probability measures the probability of an event \(A\) occurring given that another event \(B\) has already occurred.
The formula for conditional probability is:
\( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}, \quad \text{where } P(B) \neq 0 \)
If events \(A\) and \(B\) are independent, then:
\( P(A \mid B) = P(A) \)
- Tree diagrams and sample spaces can be used to visualize conditional probability problems.
- Example situations include:
- Drawing cards from a deck without replacement.
- Probabilities involving dice rolls where one condition is already satisfied.
Example
A die is rolled once. Let event A = “the number is even”, event B = “the number is greater than 3”. Find \( P(A \mid B) \).
▶️ Answer/Explanation
Sample space \( S = \{1, 2, 3, 4, 5, 6\} \)
\( A = \{2, 4, 6\}, \; B = \{4, 5, 6\} \)
\( A \cap B = \{4, 6\} \)
\( P(A \cap B) = \dfrac{2}{6} = \dfrac{1}{3}, \quad P(B) = \dfrac{3}{6} = \dfrac{1}{2} \)
\( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{\tfrac{1}{3}}{\tfrac{1}{2}} = \dfrac{2}{3} \)
Final Answer: \(\boxed{\dfrac{2}{3}}\)
Example
A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement. Find \( P(\text{second red} \mid \text{first red}) \).
▶️ Answer/Explanation
If the first ball is red, 2 red and 2 blue remain.
\( P(\text{second red} \mid \text{first red}) = \dfrac{2}{4} = \dfrac{1}{2} \)
Final Answer: \(\boxed{\dfrac{1}{2}}\)
Example
A bag has 4 red and 1 blue ball. Two balls are drawn with replacement. Find \( P(\text{both red} \mid \text{first red}) \).
▶️ Answer/Explanation
\( P(R \text{ on first draw}) = \dfrac{4}{5} \)
Since replacement, probabilities stay the same.
\( P(\text{second R} \mid \text{first R}) = \dfrac{4}{5} \)
Therefore, \( P(\text{both R} \mid \text{first R}) = \dfrac{4}{5} \)
Final Answer: \(\boxed{\dfrac{4}{5}}\)
Example
A card is drawn at random from a standard deck of 52 cards. Let \( A = \{\text{the card is a king}\} \), \( B = \{\text{the card is a face card (J, Q, K)}\} \). Find \( P(A \mid B) \).
▶️ Answer/Explanation
\( A = 4 \ \text{kings}, \quad B = 12 \ \text{face cards (4 J, 4 Q, 4 K)} \)
\( A \cap B = 4 \ \text{kings} \)
\( P(A \cap B) = \dfrac{4}{52} = \dfrac{1}{13} \)
\( P(B) = \dfrac{12}{52} = \dfrac{3}{13} \)
\( P(A \mid B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{\tfrac{1}{13}}{\tfrac{3}{13}} = \dfrac{1}{3} \)
Final Answer: \(\boxed{\dfrac{1}{3}}\)