10.3 Tree Diagrams and Conditional Probability
Understanding Tree Diagrams
Tree diagrams show all possible outcomes of multiple events. They're especially useful when probabilities change between events. NOT TO BE CONFUSED WITH FREQUENCY TREES! Tree diagrams are used to calculate probability whereas frequency trees organise numbered data.
Tree diagrams show all possible outcomes of multiple events. They're especially useful when probabilities change between events. NOT TO BE CONFUSED WITH FREQUENCY TREES! Tree diagrams are used to calculate probability whereas frequency trees organise numbered data.
⚡ How Tree Diagrams Work:
Key Rules:
• Each branch shows one outcome
• Write probabilities on the branches
• All branches from one point must add to 1
• Follow a path to find combined probability
• Multiply along branches to find probability of a path
• Add probabilities of different paths for "or" scenarios
• Each branch shows one outcome
• Write probabilities on the branches
• All branches from one point must add to 1
• Follow a path to find combined probability
• Multiply along branches to find probability of a path
• Add probabilities of different paths for "or" scenarios
Example 1: Flipping Two Coins
Draw a tree diagram for flipping two fair coins and find P(exactly one head).
Outcomes:
• HH: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• HT: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ ✓
• TH: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ ✓
• TT: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
P(exactly one head):
P(HT or TH) = $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
Answer: $\frac{1}{2}$ or 0.5
• HH: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• HT: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ ✓
• TH: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ ✓
• TT: $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
P(exactly one head):
P(HT or TH) = $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$
Answer: $\frac{1}{2}$ or 0.5
Example 2: With Replacement
A bag contains 3 red balls and 2 blue balls. Two balls are drawn WITH replacement. Find P(both red).
With replacement means: The first ball is put back, so probabilities stay the same
P(both red) = P(RR):
Multiply along the RR path:
$$P(RR) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$$
Answer: $\frac{9}{25}$ or 0.36
With replacement means: The first ball is put back, so probabilities stay the same
Multiply along the RR path:
$$P(RR) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$$
Answer: $\frac{9}{25}$ or 0.36
Example 3: Without Replacement
A bag contains 3 red balls and 2 blue balls. Two balls are drawn WITHOUT replacement. Find P(one of each color).
Without replacement means: The first ball is NOT put back, so probabilities change
Tree Diagram:
First draw: 3R and 2B out of 5 total
Second draw: Only 4 balls left
If first is Red (3/5):
• Second Red: 2/4 (2 red left from 4 total)
• Second Blue: 2/4 (still 2 blue from 4 total)
If first is Blue (2/5):
• Second Red: 3/4 (still 3 red from 4 total)
• Second Blue: 1/4 (only 1 blue left from 4 total)
P(one of each):
P(RB or BR) = P(RB) + P(BR)
$$= \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}$$
$$= \frac{6}{20} + \frac{6}{20} = \frac{12}{20} = \frac{3}{5}$$
Answer: $\frac{3}{5}$ or 0.6
Without replacement means: The first ball is NOT put back, so probabilities change
Tree Diagram:
First draw: 3R and 2B out of 5 total
Second draw: Only 4 balls left
If first is Red (3/5):
• Second Red: 2/4 (2 red left from 4 total)
• Second Blue: 2/4 (still 2 blue from 4 total)
If first is Blue (2/5):
• Second Red: 3/4 (still 3 red from 4 total)
• Second Blue: 1/4 (only 1 blue left from 4 total)
P(one of each):
P(RB or BR) = P(RB) + P(BR)
$$= \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4}$$
$$= \frac{6}{20} + \frac{6}{20} = \frac{12}{20} = \frac{3}{5}$$
Answer: $\frac{3}{5}$ or 0.6
🎯 Tree Diagram Practice
Understanding Conditional Probability
Conditional probability is the probability of an event given that another event has already occurred. Written as P(A|B) meaning "probability of A given B".
Conditional probability is the probability of an event given that another event has already occurred. Written as P(A|B) meaning "probability of A given B".
⚡ Conditional Probability Formula:
P(A|B)
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
"Probability of A given B"
Rearranged
$$P(A \cap B) = P(A|B) \times P(B)$$
Probability of both A and B
What it means:
P(A|B) asks: "If we know B has happened, what's the probability of A?"
Example: P(rainy|cloudy) = probability it rains, given it's already cloudy
P(A|B) asks: "If we know B has happened, what's the probability of A?"
Example: P(rainy|cloudy) = probability it rains, given it's already cloudy
Example 1: Using the Formula
In a class, 60% of students passed (P), and 40% of students both passed and play sports (P ∩ S). Find P(S|P) - the probability a student plays sports given they passed.
Given:
P(P) = 0.6
P(P ∩ S) = 0.4
Formula:
$$P(S|P) = \frac{P(P \cap S)}{P(P)}$$
Solution:
$$P(S|P) = \frac{0.4}{0.6} = \frac{2}{3} \approx 0.667$$
Answer: $\frac{2}{3}$ or about 67%
Interpretation: Of students who passed, about 67% play sports
Given:
P(P) = 0.6
P(P ∩ S) = 0.4
Formula:
$$P(S|P) = \frac{P(P \cap S)}{P(P)}$$
Solution:
$$P(S|P) = \frac{0.4}{0.6} = \frac{2}{3} \approx 0.667$$
Answer: $\frac{2}{3}$ or about 67%
Interpretation: Of students who passed, about 67% play sports
Example 2: Two-Way Table
A survey of 100 people asked about pet ownership:
Find P(has cat | has dog):
This means: "Of people who have dogs, what proportion have cats?"
Of the 50 people with dogs, 15 also have cats
$$P(\text{cat}|\text{dog}) = \frac{15}{50} = \frac{3}{10} = 0.3$$
Answer: 0.3 or 30%
| Has Dog | No Dog | Total | |
|---|---|---|---|
| Has Cat | 15 | 25 | 40 |
| No Cat | 35 | 25 | 60 |
| Total | 50 | 50 | 100 |
This means: "Of people who have dogs, what proportion have cats?"
Of the 50 people with dogs, 15 also have cats
$$P(\text{cat}|\text{dog}) = \frac{15}{50} = \frac{3}{10} = 0.3$$
Answer: 0.3 or 30%
Example 3: Medical Testing
A disease affects 2% of the population. A test for the disease is 95% accurate (correct 95% of the time for both positive and negative cases).
Find: P(has disease | positive test)
Given:
• P(disease) = 0.02
• P(positive | disease) = 0.95
• P(negative | no disease) = 0.95, so P(positive | no disease) = 0.05
Using tree diagram logic:
P(disease AND positive) = 0.02 × 0.95 = 0.019
P(no disease AND positive) = 0.98 × 0.05 = 0.049
P(positive) = 0.019 + 0.049 = 0.068
$$P(\text{disease}|\text{positive}) = \frac{0.019}{0.068} = 0.279$$
Answer: About 28%
Surprising result: Even with a positive test, there's only a 28% chance of having the disease! This is because the disease is rare.
Find: P(has disease | positive test)
Given:
• P(disease) = 0.02
• P(positive | disease) = 0.95
• P(negative | no disease) = 0.95, so P(positive | no disease) = 0.05
Using tree diagram logic:
P(disease AND positive) = 0.02 × 0.95 = 0.019
P(no disease AND positive) = 0.98 × 0.05 = 0.049
P(positive) = 0.019 + 0.049 = 0.068
$$P(\text{disease}|\text{positive}) = \frac{0.019}{0.068} = 0.279$$
Answer: About 28%
Surprising result: Even with a positive test, there's only a 28% chance of having the disease! This is because the disease is rare.
💡 Conditional Probability Tips:
• The vertical bar "|" means "given": What we know has happened
• Focus on the condition: Only consider cases where B occurred
• From tables: Look at the row/column for the condition
• Order matters: P(A|B) ≠ P(B|A) in general
• Tree diagrams help: Follow the branch for the condition
• Focus on the condition: Only consider cases where B occurred
• From tables: Look at the row/column for the condition
• Order matters: P(A|B) ≠ P(B|A) in general
• Tree diagrams help: Follow the branch for the condition
🎯 Conditional Probability Practice:
Combining Probabilities
The AND rule and OR rule help us calculate probabilities of combined events.
The AND rule and OR rule help us calculate probabilities of combined events.
⚡ The Two Rules:
AND Rule (Multiply)
P(A and B) = P(A) × P(B|A)
For independent events:
P(A and B) = P(A) × P(B)
OR Rule (Add)
P(A or B) = P(A) + P(B) - P(A and B)
For mutually exclusive:
P(A or B) = P(A) + P(B)
Key Terms:
• Independent: One event doesn't affect the other (coin flips)
• Mutually exclusive: Can't happen at the same time (rolling 2 and 3)
• AND means multiply (along branches)
• OR means add (different paths)
• Independent: One event doesn't affect the other (coin flips)
• Mutually exclusive: Can't happen at the same time (rolling 2 and 3)
• AND means multiply (along branches)
• OR means add (different paths)
Example 1: Independent Events (AND)
The probability of rain tomorrow is 0.3. The probability of traffic is 0.4. Assuming these are independent, find P(rain AND traffic).
Since independent:
$$P(\text{rain AND traffic}) = P(\text{rain}) \times P(\text{traffic})$$
$$= 0.3 \times 0.4 = 0.12$$
Answer: 0.12 or 12%
Since independent:
$$P(\text{rain AND traffic}) = P(\text{rain}) \times P(\text{traffic})$$
$$= 0.3 \times 0.4 = 0.12$$
Answer: 0.12 or 12%
Example 2: Mutually Exclusive Events (OR)
When rolling a die, find P(rolling 2 OR 5).
These are mutually exclusive (can't roll both at once)
$$P(2 \text{ OR } 5) = P(2) + P(5)$$
$$= \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$
Answer: $\frac{1}{3}$ or about 0.333
These are mutually exclusive (can't roll both at once)
$$P(2 \text{ OR } 5) = P(2) + P(5)$$
$$= \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$$
Answer: $\frac{1}{3}$ or about 0.333
Example 3: Non-Mutually Exclusive (OR)
In a deck of cards, find P(drawing a heart OR a king).
These are NOT mutually exclusive (King of Hearts is both)
Use full OR rule:
$$P(\text{heart OR king}) = P(\text{heart}) + P(\text{king}) - P(\text{heart AND king})$$
$$= \frac{13}{52} + \frac{4}{52} - \frac{1}{52}$$
$$= \frac{16}{52} = \frac{4}{13}$$
Answer: $\frac{4}{13}$ or about 0.308
Why subtract? We counted the King of Hearts twice
These are NOT mutually exclusive (King of Hearts is both)
Use full OR rule:
$$P(\text{heart OR king}) = P(\text{heart}) + P(\text{king}) - P(\text{heart AND king})$$
$$= \frac{13}{52} + \frac{4}{52} - \frac{1}{52}$$
$$= \frac{16}{52} = \frac{4}{13}$$
Answer: $\frac{4}{13}$ or about 0.308
Why subtract? We counted the King of Hearts twice
Example 4: Combined Rules
A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find P(at least one red).
Method 1: Use complement
P(at least one red) = 1 - P(both blue)
$$P(\text{both blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$$
$$P(\text{at least one red}) = 1 - \frac{3}{28} = \frac{25}{28}$$
Method 2: Add all cases with red
P(RR) + P(RB) + P(BR)
$$= \frac{5}{8} \times \frac{4}{7} + \frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7}$$
$$= \frac{20}{56} + \frac{15}{56} + \frac{15}{56} = \frac{50}{56} = \frac{25}{28}$$
Answer: $\frac{25}{28}$ or about 0.893
Method 1: Use complement
P(at least one red) = 1 - P(both blue)
$$P(\text{both blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$$
$$P(\text{at least one red}) = 1 - \frac{3}{28} = \frac{25}{28}$$
Method 2: Add all cases with red
P(RR) + P(RB) + P(BR)
$$= \frac{5}{8} \times \frac{4}{7} + \frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7}$$
$$= \frac{20}{56} + \frac{15}{56} + \frac{15}{56} = \frac{50}{56} = \frac{25}{28}$$
Answer: $\frac{25}{28}$ or about 0.893
💡 AND/OR Rule Tips:
• AND = multiply: Both must happen
• OR = add (carefully): At least one happens
• Check for overlap: If OR events can happen together, subtract the overlap
• "At least one": Use 1 - P(none) - it's usually easier!
• Independent check: Does one affect the other?
• OR = add (carefully): At least one happens
• Check for overlap: If OR events can happen together, subtract the overlap
• "At least one": Use 1 - P(none) - it's usually easier!
• Independent check: Does one affect the other?
🎯 AND/OR Practice
Real Life Uses:
• Insurance: Calculating risk of multiple events
• Medical diagnosis: Combining test results
• Quality control: Probability of defect-free products
• Weather forecasting: Combined probabilities
• Sports betting: Multiple game outcomes
• Network reliability: System functioning probabilities
• Insurance: Calculating risk of multiple events
• Medical diagnosis: Combining test results
• Quality control: Probability of defect-free products
• Weather forecasting: Combined probabilities
• Sports betting: Multiple game outcomes
• Network reliability: System functioning probabilities