IB MYP 4-5 Maths- Addition and multiplication rule Conditional Probability- Study Notes - New Syllabus
IB MYP 4-5 Maths- Addition and multiplication rule-conditional probability – Study Notes
Extended
- Addition and multiplication rule-conditional probability
Conditional Probability
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred.
It is denoted as \( P(A|B) \), which means “the probability of event A given that event B has occurred”.
Formula:
\( P(A|B) = \dfrac{P(A \cap B)}{P(B)} \), where \( P(B) > 0 \).
Interpretation:
- If \( P(A|B) > P(A) \): Event B increases the likelihood of A.
- If \( P(A|B) = P(A) \): Events A and B are independent.
Key Properties:
- \( P(A \cap B) = P(B) \cdot P(A|B) = P(A) \cdot P(B|A) \)
- If A and B are independent, then \( P(A|B) = P(A) \) and \( P(B|A) = P(B) \).
Addition Rule in Conditional Probability
For any two events A and B:
\( P(A \cup B) = P(A) + P(B) – P(A \cap B) \)
If events are mutually exclusive: \( P(A \cup B) = P(A) + P(B) \).
Multiplication Rule in Conditional Probability
\( P(A \cap B) = P(A|B) \cdot P(B) = P(B|A) \cdot P(A) \)
Example:
A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability that the second ball is red, given that the first ball is red.
▶️Answer/Explanation
Step 1: Total balls = 8.
Step 2: After first red is drawn, remaining balls = 7, red balls = 4.
\( P(\text{Second red | First red}) = \dfrac{\text{Remaining red balls}}{\text{Remaining total}} = \dfrac{4}{7} \).
Answer: \( \dfrac{4}{7} \).
Example:
A student is selected randomly. The probability that the student studies Math is 0.6, and English is 0.5. The probability the student studies both is 0.3. Find the probability that the student studies Math or English.
▶️Answer/Explanation
\( P(M \cup E) = P(M) + P(E) – P(M \cap E) \)
\( = 0.6 + 0.5 – 0.3 = 0.8 \)
Answer: \( 0.8 \).
Example:
The probability that a person likes tea is 0.5, coffee is 0.4, and both tea and coffee is 0.2. Find \( P(\text{Tea and Coffee}) \) using multiplication rule.
▶️Answer/Explanation
\( P(T \cap C) = P(T) \cdot P(C|T) \)
\( 0.2 = 0.5 \cdot P(C|T) \Rightarrow P(C|T) = \dfrac{0.2}{0.5} = 0.4 \)
Interpretation: Probability of coffee given tea = 0.4.
Example:
A survey was conducted among 100 students about whether they like Tea or Coffee:
- 40 like Tea
- 50 like Coffee
- 20 like both Tea and Coffee
Find the probability that a student likes Tea given that the student likes Coffee.
▶️Answer/Explanation
Step 1: Construct a two-way table:
| Coffee | Not Coffee | Total | |
|---|---|---|---|
| Tea | 20 | 20 | 40 |
| Not Tea | 30 | 30 | 60 |
| Total | 50 | 50 | 100 |
Step 2: We need \( P(\text{Tea | Coffee}) = \dfrac{\text{Tea and Coffee}}{\text{Coffee}} \)
\( = \dfrac{20}{50} = 0.4 \)
Answer: \( 0.4 \) (or 40%).
Example :
Suppose there are three sports teams at school:
- Football (F)
- Basketball (B)
- Track (T)
Given:
- 15 students play Football (F)
- 12 play Basketball (B)
- 10 run Track (T)
- 8 play Football and Basketball
- 4 play Football and Track
- 4 play Basketball and Track
- 3 students play all three sports
Find:
- \( P(\text{Football and Basketball}) \)
- \( P(\text{Football or Basketball}) \)
- \( P(\text{Track | Football}) \)
▶️Answer/Explanation
Step : Break down with Venn diagram values:
- \( F \cap B \cap T = 3 \)
- \( F \cap B = 8 \) → so only F & B = \( 8 – 3 = 5 \)
- \( F \cap T = 4 \) → so only F & T = \( 4 – 3 = 1 \)
- \( B \cap T = 4 \) → so only B & T = \( 4 – 3 = 1 \)
Now, only Football = \( 15 – (5 + 1 + 3) = 6 \)
Only Basketball = \( 12 – (5 + 1 + 3) = 3 \)
Only Track = \( 10 – (1 + 1 + 3) = 5 \)
Step : Total students in at least one sport:
\( 6 + 3 + 5 + 5 + 1 + 1 + 3 = 24 \)
(i) Football and Basketball:
\( P(F \cap B) = \dfrac{8}{24} = \dfrac{1}{3} \)
(ii) Football or Basketball:
\( P(F \cup B) = P(F) + P(B) – P(F \cap B) \)
\( P(F) = \dfrac{15}{24}, \; P(B) = \dfrac{12}{24}, \; P(F \cap B) = \dfrac{8}{24} \)
So, \( P(F \cup B) = \dfrac{15 + 12 – 8}{24} = \dfrac{19}{24} \)
(iii) Track given Football:
\( P(T | F) = \dfrac{P(F \cap T)}{P(F)} = \dfrac{4}{15} \)
Important Notes:
- Conditional probability is useful when events are dependent.
- Tree diagrams and tables are often used to illustrate conditional probabilities.
- Multiplication rule is essential for joint probability.