4.7 Conditional Probability and Independent Events

1. Conditional Probability

Notice the difference in notation:

  • $P(A)$ means "probability of $A$"
  • $P(A|B)$ means "probability of $A$, given $B$"

The probability that it will rain on a given day differs from the probability that it will rain given that the day is in September.

For example, suppose we pick an integer from 1 to 100. Let $A = \text{"we pick 17"}$. Then $P(A) = \dfrac{1}{100}$. If we know $B = \text{"the number has two digits"}$, then $P(A|B) = \dfrac{1}{90}$ because there are 90 two-digit numbers in that range.

Definition of $P(A|B)$

Conditional probability is given by the formula:

$P(A|B) = \dfrac{n(A \cap B)}{n(B)} \quad \text{or} \quad P(A|B) = \dfrac{P(A \cap B)}{P(B)}$

Venn diagrams and tables help clarify this definition.

EXAMPLE 1: $P(A|B)$ in a Venn Diagram

S 100 A B 20 10 30 40
From the diagram, $P(A) = \dfrac{30}{100}$. To find $P(A|B)$, restrict the sample space to event $B$. The total count is now 40 instead of 100:
$P(A|B) = \dfrac{?}{40} \longleftarrow \text{given } B$
There are 10 elements of $A$ inside $B$. Therefore:
$P(A|B) = \dfrac{10}{40} = 0.25$
This applies the formula $P(A|B) = \dfrac{n(A \cap B)}{n(B)} = \dfrac{10}{40} = 0.25$.
Dividing the numerator and denominator by the total sample space size yields the formal definition:
$P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{\dfrac{10}{100}}{\dfrac{40}{100}} = \mathbf{0.25}$
Similarly:
  • $P(B|A) = \dfrac{10}{30} \longleftarrow \text{given } A$
  • $P(A'|B) = \dfrac{30}{40}$
  • $P(A|B') = \dfrac{20}{60}$
  • $P(A'|B') = \dfrac{40}{60}$

EXAMPLE 2: $P(A|B)$ in a Table

Tables display conditional probabilities clearly. Consider a sample of 200 people:

Male Female Total
Smoker 40 20 60
Non-smoker 80 60 140
Total 120 80 200

Compare standard and conditional probabilities:

  • $P(\text{smoker}) = \dfrac{60}{200}$
  • $P(\text{smoker}|\text{male}) = \dfrac{40}{120} \longleftarrow \text{given male}$
  • $P(\text{male}|\text{smoker}) = \dfrac{40}{60} \longleftarrow \text{given smoker}$
  • $P(\text{female}|\text{smoker}) = \dfrac{20}{60} \approx 0.33$
  • $P(\text{non-smoker}|\text{female}) = \dfrac{60}{80} = 0.75$

2. Independent Events

Events $A$ and $B$ are independent if:

$P(A|B) = P(A)$

Event $B$ does not affect $A$. The probability of $A$ remains unchanged whether $B$ occurs or not. Consequently, $P(B|A) = P(B)$.

Substituting this into the conditional probability formula yields the multiplication rule:
$P(A \cap B) = P(A|B) \cdot P(B) \implies \mathbf{P(A \cap B) = P(A) \cdot P(B)}$

If $A$ and $B$ are independent:
(1)$P(A|B) = P(A)$
(2)$P(B|A) = P(B)$
(3)$P(A \cap B) = P(A) \cdot P(B)$

EXAMPLE 3: Proving Independence

S 120 A B 20 10 30 60
Three ways to verify independence using this diagram:
  • $P(A) = \dfrac{30}{120} = \dfrac{1}{4}$ and $P(A|B) = \dfrac{10}{40} = \dfrac{1}{4}$. Condition (1) holds.
  • $P(B) = \dfrac{40}{120} = \dfrac{1}{3}$ and $P(B|A) = \dfrac{10}{30} = \dfrac{1}{3}$. Condition (2) holds.
  • $P(A \cap B) = \dfrac{10}{120} = \dfrac{1}{12}$ and $P(A) \cdot P(B) = \dfrac{1}{4} \cdot \dfrac{1}{3} = \dfrac{1}{12}$. Condition (3) holds.

Difference: Mutually Exclusive vs. Independent

Do not confuse mutually exclusive events with independent events:

Mutually Exclusive Events: $A \cap B = \emptyset$.
This means $P(A \cap B) = 0$.
The addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ applies generally.
Independent Events: $P(A \cap B) = P(A) \cdot P(B)$.
For independent events, the addition rule simplifies to:
$P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)$
When events are known to be independent, apply the multiplication rule directly.
Example: Roll a die and flip a coin. Find the probability of rolling a six and flipping heads.
Let $A = \text{"die shows 6"}$ $\implies P(A) = \dfrac{1}{6}$.
Let $B = \text{"coin shows heads"}$ $\implies P(B) = \dfrac{1}{2}$.
These events are independent, so:
$P(A \cap B) = P(A) \cdot P(B) = \dfrac{1}{6} \cdot \dfrac{1}{2} = \mathbf{\dfrac{1}{12}}$

EXAMPLE 4

Let $P(A) = 0.4$ and $P(B) = 0.3$. Find $P(A \cup B)$ for each case:

a) $A$ and $B$ are mutually exclusive
$P(A \cup B) = P(A) + P(B) = 0.4 + 0.3 = \mathbf{0.7}$
b) $A$ and $B$ are independent
$P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B) = 0.4 + 0.3 - (0.4)(0.3) = \mathbf{0.58}$
c) Given that $P(A \cap B) = 0.2$
$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.2 = \mathbf{0.5}$
d) Given that $P(A|B) = 0.2$
Find the intersection: $P(A|B) = \dfrac{P(A \cap B)}{P(B)} \implies P(A \cap B) = P(A|B)P(B) = (0.2)(0.3) = 0.06$.
Then: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.06 = \mathbf{0.64}$

EXAMPLE 5

Let $A$ and $B$ be independent events where $P(A) = 0.4$ and $P(A \cup B) = 0.7$. Find $P(B)$.

For independent events: $P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)$
Substitute values: $\implies 0.7 = 0.4 + P(B) - 0.4P(B)$
Simplify: $\implies 0.3 = 0.6P(B)$
Solve: $\implies P(B) = \dfrac{0.3}{0.6} = \mathbf{0.5}$