A bag has 3 rust and 2 light balls. If the first ball was rust, does the chance of a second rust ball change? Conditional probability recalculates chances after known information. Replacement determines whether the bag’s composition changed. Use the probability model.
Let A mean first rust and B second rust. Each ball is distinct and selection among available balls uniform. The vertical bar means given:
P(B∣A) = (P(A∩B))/(P(A)) (P(A) > 0).
The denominator sets the new space: among A outcomes, what fraction also has B? With P(A)=0 this definition is unavailable. Rearranging gives multiplication:
P(A∩B) = P(A)P(B∣A).
With replacement
Replace the first ball and mix uniformly again. Composition is unchanged, P(B|A)=3/5=P(B), so:
P(A∩B) = (3)/(5) ⋅ (3)/(5) = (9)/(25).
This models independent draws. Replacement without adequate mixing alone does not prove that a real mechanism is independent.
Without replacement
After A, 2 rust and 2 light balls remain. The updated chance is 2/4:
P(B∣A) = (2)/(4) = (1)/(2), P(A∩B) = (3)/(5) ⋅ (1)/(2) = (3)/(10).
Two draws without replacement
Do not multiply 3/5 by 3/5 when information changed the second factor. OpenStax
The same problem in a table
Without replacement there are 5·4=20 equally likely ordered pairs of different labeled balls. The table counts possible outcomes, not 20 performed experiments:
| First / second | Rust | Light | Total |
|---|---|---|---|
| Rust | 6 | 6 | 12 |
| Light | 6 | 2 | 8 |
| Total | 12 | 8 | 20 |
Given A, only the row of 12 outcomes remains; 6 also have B: 6/12=1/2. The joint chance A∩B is instead 6/20=3/10. The overall chance B remains 12/20=3/5; an unchanged marginal chance does not imply independence from the first outcome. Contingency tables.
Independent does not mean mutually exclusive
Independence means:
P(A∩B) = P(A)P(B).
For positive P(A), this is equivalent to P(B|A)=P(B). Mutually exclusive events cannot occur together: their intersection is empty. First rust and first light are mutually exclusive; since both chances are positive, they are not independent. First rust and second rust can occur together. Definitions and distinction.
The preceding article’s addition rule also applies to dependent events. Without replacement, the chance of at least one rust ball is:
P(A∪B) = (3)/(5) + (3)/(5) − (3)/(10) = (9)/(10).
Multiply along tree paths, add mutually exclusive paths and always identify a table’s denominator. This reveals which information entered the calculation.