11-15-2025, 04:30 PM
Chapter 6 — Dependent Events
Dependent events are the first MAJOR challenge in probability.
But once you understand the idea clearly, they become simple.
This chapter explains dependent events step-by-step, using logic first, numbers second.
---
6.1 What Are Dependent Events?
Two events are dependent when:
The first event affects the second event.
This usually happens when you:
• pick something WITHOUT replacing it
• remove objects from a group
• deal cards
• choose people from a list
• take sweets from a bag
If the total changes → the probability changes.
---
6.2 The Key Rule
For dependent events:
Probability of both happening = multiply the UPDATED probabilities
Example:
A bag has 3 red and 1 blue sweet.
Pick one WITHOUT replacing it, then pick again.
P(red first) = 3/4
After removing a red, the bag has:
2 red, 1 blue → 3 sweets total
P(red second) = 2/3
So:
P(red then red) = (3/4) × (2/3) = 6/12 = 1/2
The secret is updating the second probability.
---
6.3 Why the Probability Changes
Think about this:
If you remove one item from a group,
the whole group changes.
Example:
Bag of 5 sweets:
• 2 blue
• 3 red
If you remove a red:
• now only 2 red remain
• total becomes 4
So ANY second pick depends on the first.
---
6.4 Replacement vs No Replacement — Final Clarification
With replacement → independent
(total stays the same)
Without replacement → dependent
(total changes)
Example:
Deck of cards, no replacement:
P(ace then ace)
= 4/52 × 3/51
= 1/13 × 3/51
= 3/663
Example:
Deck of cards, with replacement:
P(ace then ace)
= 4/52 × 4/52
= 1/13 × 1/13
= 1/169
Notice how the second probability is different.
---
6.5 Worked Examples
Example 1 — Sweets
A bag has 5 red and 3 blue sweets.
Pick one WITHOUT replacement, then pick again.
Find P(red then blue).
P(red first) = 5/8
After removing a red:
4 red, 3 blue → total 7
P(blue second) = 3/7
Probability = (5/8) × (3/7) = 15/56
---
Example 2 — Cards
Find P(drawing a king then a queen from a deck, no replacement).
Kings = 4
Queens = 4
Total = 52
P(king first) = 4/52 = 1/13
Then:
Queens remaining = 4
Total = 51
P(queen second) = 4/51
Multiply:
1/13 × 4/51 = 4/663
---
Example 3 — Marbles
A jar has 6 green, 2 yellow, 2 red.
Pick a marble, do NOT replace, then pick again.
Find P(green then green).
P(green first) = 6/10 = 3/5
After removing one green:
Green = 5
Total = 9
P(green second) = 5/9
Probability = (3/5) × (5/9) = 15/45 = 1/3
---
Example 4 — Students
A group has 12 boys and 8 girls.
Two students are selected without replacement.
Find P(boy then girl).
P(boy first) = 12/20 = 3/5
After removing one boy:
Boys = 11
Girls = 8
Total = 19
P(girl second) = 8/19
Probability = (3/5) × (8/19) = 24/95
---
6.6 Common Mistakes
Mistake 1: Forgetting to change the second probability
You MUST update both:
• the part
• the whole
Mistake 2: Mixing up order
P(red then blue) ≠ P(blue then red)
Mistake 3: Using replacement logic when there is no replacement
This leads to wrong denominators.
Mistake 4: Not subtracting correctly
Removing one from the total is the key.
---
6.7 Your Turn
1. A bag has 4 yellow and 6 purple beads.
Pick one without replacement.
Find P(yellow then purple).
2. A deck has 52 cards.
Find P(spade then king), no replacement.
3. A box has 3 blue, 3 red, 1 green.
Find P(red then green), no replacement.
4. In a class of 15 students, 9 are girls.
Two are chosen without replacement.
Find P(girl then girl).
5. A jar contains 7 sweets.
5 are strawberry, 2 are lemon.
Find P(lemon then strawberry), no replacement.
---
Chapter Summary
• If the total changes, the events are dependent
• The second probability must ALWAYS update
• Multiply for both independent and dependent events
• Dependent events appear in nearly every exam
• Understanding this chapter makes tree diagrams easy
---
Written and Compiled by Lee Johnston — Founder of The Lumin Archive
Dependent events are the first MAJOR challenge in probability.
But once you understand the idea clearly, they become simple.
This chapter explains dependent events step-by-step, using logic first, numbers second.
---
6.1 What Are Dependent Events?
Two events are dependent when:
The first event affects the second event.
This usually happens when you:
• pick something WITHOUT replacing it
• remove objects from a group
• deal cards
• choose people from a list
• take sweets from a bag
If the total changes → the probability changes.
---
6.2 The Key Rule
For dependent events:
Probability of both happening = multiply the UPDATED probabilities
Example:
A bag has 3 red and 1 blue sweet.
Pick one WITHOUT replacing it, then pick again.
P(red first) = 3/4
After removing a red, the bag has:
2 red, 1 blue → 3 sweets total
P(red second) = 2/3
So:
P(red then red) = (3/4) × (2/3) = 6/12 = 1/2
The secret is updating the second probability.
---
6.3 Why the Probability Changes
Think about this:
If you remove one item from a group,
the whole group changes.
Example:
Bag of 5 sweets:
• 2 blue
• 3 red
If you remove a red:
• now only 2 red remain
• total becomes 4
So ANY second pick depends on the first.
---
6.4 Replacement vs No Replacement — Final Clarification
With replacement → independent
(total stays the same)
Without replacement → dependent
(total changes)
Example:
Deck of cards, no replacement:
P(ace then ace)
= 4/52 × 3/51
= 1/13 × 3/51
= 3/663
Example:
Deck of cards, with replacement:
P(ace then ace)
= 4/52 × 4/52
= 1/13 × 1/13
= 1/169
Notice how the second probability is different.
---
6.5 Worked Examples
Example 1 — Sweets
A bag has 5 red and 3 blue sweets.
Pick one WITHOUT replacement, then pick again.
Find P(red then blue).
P(red first) = 5/8
After removing a red:
4 red, 3 blue → total 7
P(blue second) = 3/7
Probability = (5/8) × (3/7) = 15/56
---
Example 2 — Cards
Find P(drawing a king then a queen from a deck, no replacement).
Kings = 4
Queens = 4
Total = 52
P(king first) = 4/52 = 1/13
Then:
Queens remaining = 4
Total = 51
P(queen second) = 4/51
Multiply:
1/13 × 4/51 = 4/663
---
Example 3 — Marbles
A jar has 6 green, 2 yellow, 2 red.
Pick a marble, do NOT replace, then pick again.
Find P(green then green).
P(green first) = 6/10 = 3/5
After removing one green:
Green = 5
Total = 9
P(green second) = 5/9
Probability = (3/5) × (5/9) = 15/45 = 1/3
---
Example 4 — Students
A group has 12 boys and 8 girls.
Two students are selected without replacement.
Find P(boy then girl).
P(boy first) = 12/20 = 3/5
After removing one boy:
Boys = 11
Girls = 8
Total = 19
P(girl second) = 8/19
Probability = (3/5) × (8/19) = 24/95
---
6.6 Common Mistakes
Mistake 1: Forgetting to change the second probability
You MUST update both:
• the part
• the whole
Mistake 2: Mixing up order
P(red then blue) ≠ P(blue then red)
Mistake 3: Using replacement logic when there is no replacement
This leads to wrong denominators.
Mistake 4: Not subtracting correctly
Removing one from the total is the key.
---
6.7 Your Turn
1. A bag has 4 yellow and 6 purple beads.
Pick one without replacement.
Find P(yellow then purple).
2. A deck has 52 cards.
Find P(spade then king), no replacement.
3. A box has 3 blue, 3 red, 1 green.
Find P(red then green), no replacement.
4. In a class of 15 students, 9 are girls.
Two are chosen without replacement.
Find P(girl then girl).
5. A jar contains 7 sweets.
5 are strawberry, 2 are lemon.
Find P(lemon then strawberry), no replacement.
---
Chapter Summary
• If the total changes, the events are dependent
• The second probability must ALWAYS update
• Multiply for both independent and dependent events
• Dependent events appear in nearly every exam
• Understanding this chapter makes tree diagrams easy
---
Written and Compiled by Lee Johnston — Founder of The Lumin Archive