Addition rule
Why subtract intersection?: If events overlap, we count intersection twice.
Subtract once to correct.
Worked example
P(A)=0.4, P(B)=0.3, P(both)=0.1.
Find P(A or B).
Solution
- P(A or B)=0.4+0.3-0.1
- P(A or B)=0.6
Final answer
0.6 (subtract 0.1 to avoid double-counting).
Multiplication rule and independence
Independent events: Events A and B independent if P(B|A)=P(B).
Then: P(A and B)=P(A)×P(B).
Worked example
Roll die, flip coin.
P(3 and H)?
Solution
- P(3)=1/6, P(H)=1/2 (independent)
- P(3 and H)=(1/6)×(1/2)=1/12
Final answer
1/12. Independent so multiply directly.
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Conditional probability
Meaning: Probability of B given A happened.
Restricts sample space to outcomes where A occurs.
Worked example
Cards: 5 red, 3 black.
Pick one.
If red, prob of second red?
(no replacement)
Solution
- Given first is red: 4 red left, 7 total left
- P(second red | first red)=4/7
Final answer
4/7. Sample space shrinks after first pick.
Organizing combined events
Tree diagrams: Show all paths.
Label branches with probabilities.
Final outcomes: multiply along path.
Probability tables: Organize outcomes in rows and columns.
Easier for many events.
Worked example
Bag: 3 red, 2 blue.
Draw 2 (no replace).
Find all P outcomes.
Solution
- First red (3/5): second red (2/4)=6/20
- First red (3/5): second blue (2/4)=6/20
- First blue (2/5): second red (3/4)=6/20
- First blue (2/5): second blue (1/4)=2/20
Final answer
RR=6/20, RB=6/20, BR=6/20, BB=2/20. Total=20/20=1.
IB-style question — probability from a two-way table
Of 100 households: 40 own a dog, and 25 of those also own a cat. Of the 60 without a dog, 10 own a cat.
Find (a) P(owns a cat) and (b) P(owns a cat | owns a dog).
Step by step
- Fill the table: dog&cat = 25, dog-only = 15, cat-only = 10, neither = 50. Total cats = 25 + 10 = 35.
- (b) Conditional — restrict to the 40 dog owners.
Final answer
(a) 0.35. (b) 0.625.