Solved Exercises on Conditional Probability in Grade 10

Master conditional probability: Bayes theorem, tree diagrams, independence, and probability rules through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Basic Conditional Probability
Exercise 1
A bag contains 5 red balls and 3 blue balls. Two balls are drawn without replacement. Find the probability that the second ball is red given that the first ball was red.
Definition:

Conditional Probability: P(A|B) = P(A and B)/P(B), probability of A given B occurred

Without replacement: First draw affects probability of second draw

Dependent events: Outcome of first event affects second event

Conditional probability method:
  1. Identify the condition (given event)
  2. Determine the reduced sample space after condition
  3. Count favorable outcomes in reduced space
  4. Calculate probability
Initial
5R, 3B (8 total)
After 1st red
4R, 3B (7 total)
P(2nd red|1st red)
4/7
Step 1: Identify the condition

Given: First ball is red

Step 2: Update sample space

Original: 5 red, 3 blue = 8 balls

After removing 1 red: 4 red, 3 blue = 7 balls

Step 3: Count favorable outcomes

Number of red balls remaining = 4

Step 4: Calculate probability

P(2nd red | 1st red) = Number of red balls remaining / Total balls remaining

P(2nd red | 1st red) = 4/7

P(2nd red | 1st red) = 4/7
Final answer:

The probability that the second ball is red given that the first ball was red is 4/7

Applied rules:

• Conditional probability: P(A|B) = P(A and B)/P(B)

• Without replacement: Sample space reduces after each draw

• Dependence: Events are dependent when replacement doesn't occur

2 Tree Diagram
Exercise 2
In a company, 60% of employees are female. Among females, 30% have management positions. Among males, 40% have management positions. Find the probability that a randomly selected employee is female given they have a management position.
Definition:

Tree Diagram: Visual representation of sequential probability events

Bayes Theorem: P(A|B) = P(B|A)·P(A) / P(B)

Total Probability: P(B) = P(B|A)·P(A) + P(B|A')·P(A')

Given
P(F)=0.6, P(M|F)=0.3, P(M|M')=0.4
Bayes calc
P(F|M) = P(M|F)·P(F)/P(M)
Result
0.529
Step 1: Define events

F = employee is female, M = employee has management position

P(F) = 0.6, P(F') = 0.4 (male)

P(M|F) = 0.3, P(M|F') = 0.4

Step 2: Apply total probability rule

P(M) = P(M|F)·P(F) + P(M|F')·P(F')

P(M) = 0.3×0.6 + 0.4×0.4 = 0.18 + 0.16 = 0.34

Step 3: Apply Bayes theorem

P(F|M) = P(M|F)·P(F) / P(M)

P(F|M) = (0.3×0.6) / 0.34 = 0.18 / 0.34 = 0.529

Step 4: Interpret result

There is a 52.9% chance that a management employee is female

P(Female | Management) = 0.529
Final answer:

The probability that a randomly selected employee is female given they have a management position is approximately 0.529 or 52.9%

Applied rules:

• Bayes theorem: P(A|B) = P(B|A)·P(A) / P(B)

• Total probability: P(B) = Σ P(B|Ai)·P(Ai)

• Tree diagram: Multiply along branches, add across outcomes

3 Independence
Exercise 3
Two events A and B have probabilities P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.2. Are A and B independent? Justify your answer.
Definition:

Independent Events: P(A and B) = P(A) × P(B)

Alternative test: P(A|B) = P(A) or P(B|A) = P(B)

Dependent events: Occurrence of one affects probability of other

Given
P(A)=0.4, P(B)=0.5, P(A∩B)=0.2
Independence test
P(A)×P(B) = 0.2
Compare
P(A∩B) = P(A)×P(B)
Step 1: State independence criterion

Events A and B are independent if P(A and B) = P(A) × P(B)

Step 2: Calculate P(A) × P(B)

P(A) × P(B) = 0.4 × 0.5 = 0.2

Step 3: Compare with P(A and B)

P(A and B) = 0.2

P(A) × P(B) = 0.2

Step 4: Make conclusion

Since P(A and B) = P(A) × P(B), events A and B are independent

A and B are independent
Final answer:

Yes, events A and B are independent because P(A and B) = P(A) × P(B) = 0.2

Applied rules:

• Independence test: P(A and B) = P(A) × P(B)

• Conditional probability: P(A|B) = P(A and B)/P(B)

• For independent events: P(A|B) = P(A)

Conditional Probability Rules
\(P(A|B) = \frac{P(A \cap B)}{P(B)}, P(A \cap B) = P(A) \cdot P(B|A), P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
Key Probability Formulas
Conditional Probability
P(A|B) = P(A∩B)/P(B)
Probability of A given B
Multiplication Rule
P(A∩B) = P(A)·P(B|A)
Joint probability
Independence
P(A∩B) = P(A)·P(B)
When events are independent
Key definitions:

Conditional Probability: Probability of an event given that another event has occurred

Independent Events: Events where occurrence of one doesn't affect probability of the other

Dependent Events: Events where occurrence of one affects probability of the other

Conditional probability methodology:
  1. Identify events: Define A and B clearly
  2. Determine dependence: Check if events are independent or dependent
  3. Select formula: Choose appropriate probability rule
  4. Calculate: Apply the formula systematically
  5. Interpret: State the answer in context
Tip 1: Always check if events are independent before assuming P(A∩B) = P(A)×P(B).
Tip 2: Tree diagrams help visualize sequential probability problems.
Tip 3: Bayes theorem is useful when you know P(B|A) but need P(A|B).
Tip 4: Remember P(A|B) ≠ P(B|A) unless P(A) = P(B).
Common errors: Confusing P(A|B) with P(B|A), assuming independence when not applicable, calculation mistakes.
Exam preparation: Practice tree diagrams, memorize Bayes theorem, work with real-world scenarios.
Essential formulas:

• Conditional probability: P(A|B) = P(A∩B)/P(B)

• Multiplication rule: P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B)

• Independence: P(A∩B) = P(A)·P(B)

• Addition rule: P(A∪B) = P(A) + P(B) - P(A∩B)

• Bayes theorem: P(A|B) = P(B|A)·P(A)/P(B)

Solution: Exercises 4 to 5
4 Medical Testing
Exercise 4
A disease affects 2% of the population. A test is 95% accurate for those with the disease and 90% accurate for those without. Find the probability that a person has the disease given they tested positive.
Definition:

Sensitivity: P(positive test | disease) = 0.95

Specificity: P(negative test | no disease) = 0.90

Prevalence: P(disease) = 0.02

Positive Predictive Value: P(disease | positive test)

Given
P(D)=0.02, P(+|D)=0.95, P(-|D')=0.90
Bayes calc
P(D|+) = P(+|D)·P(D)/P(+)
Result
0.161
Step 1: Define events and given probabilities

D = has disease, + = positive test

P(D) = 0.02 (prevalence), P(D') = 0.98

P(+|D) = 0.95 (sensitivity), P(-|D') = 0.90 (specificity)

Therefore: P(+|D') = 1 - 0.90 = 0.10 (false positive rate)

Step 2: Apply total probability rule

P(+) = P(+|D)·P(D) + P(+|D')·P(D')

P(+) = 0.95×0.02 + 0.10×0.98 = 0.019 + 0.098 = 0.117

Step 3: Apply Bayes theorem

P(D|+) = P(+|D)·P(D) / P(+)

P(D|+) = (0.95×0.02) / 0.117 = 0.019 / 0.117 = 0.162

Step 4: Interpret result

Only about 16.2% of positive tests indicate actual disease presence

P(Disease | Positive) = 0.162
Final answer:

The probability that a person has the disease given they tested positive is approximately 0.162 or 16.2%

Applied rules:

• Bayes theorem: P(A|B) = P(B|A)·P(A)/P(B)

• Total probability: P(B) = P(B|A)·P(A) + P(B|A')·P(A')

• Medical testing: Sensitivity, specificity, prevalence relationships

5 Card Probability
Exercise 5
Two cards are drawn from a standard deck without replacement. Find the probability that both cards are face cards given that the first card is a face card.
Definition:

Face cards: Jack, Queen, King (3 per suit × 4 suits = 12 total)

Without replacement: First draw affects second draw probability

Conditional probability: P(A and B | C) = P(A and B and C)/P(C)

Given
12 face cards out of 52
After 1st face
11 face cards out of 51
P(2nd face | 1st face)
11/51
Step 1: Identify initial conditions

Standard deck: 52 cards

Face cards: J, Q, K in each of 4 suits = 3×4 = 12 face cards

Step 2: Apply the condition

Given: First card is a face card

Remaining cards: 51 total, 11 face cards

Step 3: Calculate probability of second event

P(2nd card is face | 1st card is face) = Remaining face cards / Remaining total cards

P(2nd face | 1st face) = 11/51

Step 4: Verify using conditional probability formula

P(both face | 1st face) = P(both face and 1st face) / P(1st face)

= P(both face) / P(1st face) = (12/52 × 11/51) / (12/52) = 11/51 ✓

P(2nd face | 1st face) = 11/51
Final answer:

The probability that both cards are face cards given that the first card is a face card is 11/51

Applied rules:

• Conditional probability: P(A|B) = P(A and B)/P(B)

• Without replacement: Sample space reduces after each draw

• Card counting: Track remaining cards after each draw

Detailed Summary: Conditional Probability
\(P(A|B) = \frac{P(A \cap B)}{P(B)}, P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)\)
Fundamental Conditional Probability Relations
Comprehensive definitions:

Conditional Probability: The probability of event A occurring given that event B has occurred, denoted P(A|B)

Joint Probability: Probability of both events occurring together, P(A and B) or P(A∩B)

Marginal Probability: Probability of a single event, P(A) or P(B)

Independent Events: Events where P(A|B) = P(A) and P(B|A) = P(B)

Dependent Events: Events where occurrence of one affects probability of the other

Complete conditional probability methodology:
  1. Event identification: Clearly define events A and B
  2. Condition recognition: Identify which event is given
  3. Formula selection: Choose appropriate probability formula
  4. Calculation: Apply the formula systematically
  5. Verification: Check that probability is between 0 and 1
  6. Interpretation: State the answer in context
Tip 1: Always read carefully to identify which event is the condition (the "given").
Tip 2: Tree diagrams are extremely helpful for sequential probability problems.
Tip 3: Remember that P(A|B) and P(B|A) are different quantities in general.
Tip 4: Check independence by verifying P(A and B) = P(A) × P(B).
Common applications: Medical testing, quality control, weather prediction, risk assessment.
Key relationships: P(A|B) = P(A and B)/P(B), P(A and B) = P(A)·P(B|A).
Essential formulas and rules:

• Conditional probability: P(A|B) = P(A∩B)/P(B)

• Multiplication rule: P(A∩B) = P(A)·P(B|A) = P(B)·P(A|B)

• Independence: P(A|B) = P(A), P(A∩B) = P(A)·P(B)

• Addition rule: P(A∪B) = P(A) + P(B) - P(A∩B)

• Bayes theorem: P(A|B) = P(B|A)·P(A)/P(B)

• Total probability: P(B) = P(B|A)·P(A) + P(B|A')·P(A')

Visualization: Conditional Probability Concepts
Probability Relationships
Explore conditional probability relationships:
Joint, marginal, and conditional probabilities
Independence and dependence relationships

Analysis: The chart shows different probability concepts and their relationships.

  • Conditional probability adjusts based on given information
  • Independence means conditional probability equals marginal probability
  • Joint probability represents intersection of events

Questions & Answers

Question: What's the difference between P(A|B) and P(B|A)? Aren't they the same thing?

Answer: No, P(A|B) and P(B|A) are generally NOT the same! They represent different conditional probabilities:

  • P(A|B): Probability of A given that B has occurred
  • P(B|A): Probability of B given that A has occurred

For example, if A = "person has disease" and B = "person tests positive":

  • P(B|A) = sensitivity = probability of testing positive given disease (typically high)
  • P(A|B) = positive predictive value = probability of disease given positive test (can be low even with good test)

These are related by Bayes theorem: P(A|B) = P(B|A)·P(A)/P(B)

Question: How do I know if two events are independent or dependent?

Answer: Two events A and B are independent if the occurrence of one does not affect the probability of the other. You can test this using several equivalent criteria:

  • Definition test: P(A|B) = P(A) and P(B|A) = P(B)
  • Multiplication test: P(A and B) = P(A) × P(B)
  • Practical test: Does knowing B occurred change your belief about A occurring?

Examples of independent events:

  • Flipping a coin twice
  • Drawing cards with replacement

Examples of dependent events:

  • Drawing cards without replacement
  • Weather today affecting weather tomorrow

Question: When should I use Bayes theorem versus regular conditional probability?

Answer: Use Bayes theorem when you know P(B|A) but need to find P(A|B). Regular conditional probability is used when you can directly calculate P(A and B) and P(B).

Use Bayes theorem when:

  • You have prior knowledge and want to update it based on new evidence
  • You know the "cause-effect" relationship but want the "effect-cause" probability
  • Dealing with diagnostic testing (know sensitivity/specificity, want predictive value)
  • Updating beliefs based on evidence

For example, if you know P(test positive | disease) but want P(disease | test positive), use Bayes theorem.

Regular conditional probability P(A|B) = P(A and B)/P(B) is used when you can directly determine both the numerator and denominator.

Question: Why is the probability of having a disease given a positive test result often lower than expected?

Answer: This counterintuitive result occurs due to the relationship between disease prevalence and test accuracy:

Using Bayes theorem: P(disease|positive) = [P(positive|disease) × P(disease)] / P(positive)

Where: P(positive) = P(positive|disease) × P(disease) + P(positive|no disease) × P(no disease)

Even with a highly accurate test, if the disease prevalence is low:

  • The number of true positives may be small
  • The number of false positives (from large healthy population) may be relatively large
  • This makes the ratio (true positives)/(total positives) surprisingly low

This is why screening tests for rare diseases often have low positive predictive values despite high sensitivity/specificity.

Question: How do I interpret tree diagrams for conditional probability problems?

Answer: Tree diagrams are powerful tools for conditional probability problems:

  • Each branch: Represents a possible outcome with its probability
  • Multiply along branches: To find joint probabilities
  • Add across branches: To find total probabilities of outcomes
  • Conditional probabilities: Shown on branches after the conditioning event

For example, in a two-stage process:

  • First level: Shows probabilities of initial events
  • Second level: Shows conditional probabilities given the first event
  • Path probabilities: Multiply probabilities along each path to get joint probability

Tree diagrams help organize complex sequential probability problems and make the conditional relationships clear.