14G

Conditional probability

Reduced sample spaces, the formula P(A | B) = P(A ∩ B) / P(B), the general product rule, independence, and two-way tables.

Probability "given that"

$P(A \mid B)$ is the probability of $A$ given that $B$ has happened. Knowing $B$ happened shrinks the sample space to $B$ (the reduced sample space).

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)} \qquad\text{so}\qquad P(A \cap B) = P(B) \times P(A \mid B)$$

Example 1●●

Two dice are rolled and the sum is $8$. Find the probability that it was a double.

Reduced sample space: $(2,6), (3,5), (4,4), (5,3), (6,2)$.

Only $(4,4)$ is a double, so $P = \dfrac15$.

Two-way tables

	Plays sport	No sport	Total
Year 11	30	20	50
Year 12	25	25	50
Total	55	45	100

Example 2●●

Using the table, find $P(\text{sport} \mid \text{Year 11})$ and $P(\text{Year 11} \mid \text{sport})$.

$P(\text{sport} \mid \text{Year 11}) = \dfrac{30}{50} = \dfrac35$

Restrict to the Year 11 row.

$P(\text{Year 11} \mid \text{sport}) = \dfrac{30}{55} = \dfrac{6}{11}$

Restrict to the sport column.

Independence

In the table, $P(\text{sport}) = 0.55$ but $P(\text{sport} \mid \text{Year 11}) = 0.6$, so playing sport and being in Year 11 are not independent.

Flashcards

$P(A mid B)$

$\dfrac{P(A \cap B)}{P(B)}$

General product rule

$P(A \cap B) = P(B)P(A \mid B)$

Independent

$P(A \mid B) = P(A)$

Practice

Using the table, find $P(\text{sport} \mid \text{Year 11})$.
Find $P(\text{Year 11} \mid \text{sport})$.
$P(A) = 0.5$ and $P(A \cap B) = 0.2$. Find $P(B \mid A)$.
Two dice show a sum of $8$. Find the probability it was a double.
$A$ and $B$ are independent if:
$P(B) = 0.6$ and $P(A \mid B) = 0.5$. Find $P(A \cap B)$.
Two dice are rolled and the first shows $3$. Find $P(\text{sum} = 8)$.
In the table, are "plays sport" and "Year 11" independent?
A family has two children, each equally likely to be a boy or a girl. Given at least one is a girl, find $P(\text{both girls})$.