Conditional Probability — Shrinking the Sample Space
Restricted Space — the condition shrinks U
Two dice. Let B = “first die shows 2” and A = “sum is 6”. Once B has happened, the only live outcomes are the second row.
Readout — probability relative to B:
n(B) = 6 (the restricted space — row 2)
n(A ∩ B) = 1 (the pulsing cell (2,4))
P(A|B) = n(A ∩ B) / n(B) = 1/6
Conditional probability ⭐
Working formula (frequency form) — keep the original counts: P(A|B) = n(A ∩ B) / n(B).
Example: die, U = {1, 2, 3, 4, 5, 6}, A = {4}, B = even = {2, 4, 6}; P(A|B) = n(A∩B)/n(B) = 1/3. ⭐
Multiplication Law and Independence
Multiplication Law of Probability ⭐
The probability of the simultaneous occurrence of events A and B is equal to the probability of event B multiplied by the probability of event A under the condition that event B has already occurred.
Multiplication law for independent events ⭐
Three independent events ⭐
Independence Checker
P(A) × P(B) = 0.2 vs P(A ∩ B) = 0.2
Independent ✓ — P(A ∩ B) = P(A) × P(B)
| Aspect | With replacement | Without replacement |
|---|---|---|
| 1st ball red | 5/8 | 5/8 |
| 2nd ball red (given 1st red) | 5/8 — box unchanged | 4/7 — one red removed |
| Both red | 25/64 | (5/8)(4/7) = 20/56 |
| Dependency | Draws are independent | Draws are dependent |
The Probability Tree (board workflow)
Probability Tree — 3 red + 2 blue balls, two draws
Sum of leaves: 3/10 + 3/10 + 3/10 + 1/10 = 1 ✓ — a built-in sanity check. ⭐
Without replacement the denominators shrink (2/4, 1/4) — draws are dependent. With replacement the box is unchanged (3/5, 2/5) — draws are independent.
Solved Illustrations 27–35
Solved Example
Problem
Solution
(1) 9/38 (2) 9/38 — together they make 19/38 by the addition law, a nice check.
Solved Example
Problem
Solution
(1) 1/4 (2) 1/12 (3) 11/36.
Solved Example
Problem
Solution
(1) 2/11 (2) 3/11 (3) 1/55 (4) 34/55.
Solved Example
Problem
Solution
Required probability = 28/33.
Solved Example
Problem
Solution
Required probability = 0.144.
Solved Example
Problem
Solution
(1) 0.87 (2) 0.13 (3) 0.27.
Solved Example
Problem
Solution
Required probability = 5/7.
Step 1 — the dropped book
15 of 25 books are novels; one book is dropped outside (⭐ board — drawn under a mango tree).
P(N) = 15/25 = 3/5.
Step 1 — the overlap
P(A) = 0.4 (found by A), P(B) = 0.5 (found by B), P(A ∪ B) = 0.6.
P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.4 + 0.5 − 0.6 = 0.3
Practice Checkpoint
Your turn. Work each question in your notebook — type only the final answer here.
Conditional Q13
A family has two children. If the first child is a girl, find the probability that both children in that family are girls.
Try the experiment:
Conditional Q27 (Board)
⭐ BoardFor any two events in the sample space of a random experiment, P(A) = 2/3, P(B) = 3/5 and P(B/A) = 3/4. Find the value of P(A/B).
Conditional Q31
If three events A, B and C of a random experiment are independent events and P(A) = 0.2, P(B) = 0.3 and P(C) = 0.5, find the value of P(A ∩ B ∩ C).
Key Takeaways
Key Takeaways
- Conditional probability shrinks the sample space: P(A|B) = P(A ∩ B) / P(B), P(B) ≠ 0, or by counts n(A ∩ B) / n(B).
- Multiplication law: P(A ∩ B) = P(B)·P(A|B) = P(A)·P(B|A).
- Independent events: occurrence of one does not affect the other — P(A|B) = P(A) and P(A ∩ B) = P(A)·P(B).
- Without replacement, draws are dependent (denominator shrinks); with replacement they are independent.
- The probability tree keeps both orders visible and its leaves always sum to 1 — a built-in sanity check.
- Board favourites: "exactly two of three matches" (multinomial ordering), "at least one" via the complement, and "neither" via A′ ∩ B′ = (A ∪ B)′.