Before you read

Answer first — see what you already know.

At a fitness center, 100 members were surveyed. Of these, 40 do yoga, 30 do Pilates, and 10 do both. Given that a member does yoga, what is the probability they also do Pilates?

A
$\frac{10}{100}$
B
$\frac{10}{40}$
C
$\frac{10}{30}$
D
$\frac{30}{40}$

Check your answer

Answer:

Before you read

Answer first — see what you already know.

In a clinical trial, $P (improvement) = 0.6$ , $P (treatment) = 0.5$ , and $P (improvement ∣ treatment) = 0.7$ . What is $P (improvement AND treatment)$ ?

A
0.30
B
0.35
C
0.42
D
0.50

Check your answer

Answer:

Theory

Conditional Probability

Conditional probability is the probability of $A$ given that $B$ has already occurred:
$P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$

In a two-way table, this becomes:
$P(A|B) = \frac{\text{cell (A and B)}}{\text{row or column total for B}}$

The key word is "given" — it restricts your sample space.

Probability tree for conditional events

Example 1

In a class: 60% study math, 40% study science, 25% study both. Given that a student studies math, what is the probability they also study science?

$P(S|M) = \frac{P(S \text{ and } M)}{P(M)} = \frac{0.25}{0.60} = \frac{5}{12} \approx 0.417$

\frac{5}{12} \approx 41.7\%

Theory

Independence Test

Two events are independent if knowing one doesn't change the probability of the other:
$P(A|B) = P(A)$

Equivalently: $P(A \text{ and } B) = P(A) \times P(B)$

If these equalities don't hold, the events are dependent (associated).

Example 1

Event A: $P(A) = 0.5$ . Event B: $P(B) = 0.3$ . $P(A \text{ and } B) = 0.15$ . Are A and B independent?

Check: $P(A) \times P(B) = 0.5 \times 0.3 = 0.15$

$P(A \text{ and } B) = 0.15 = P(A) \times P(B)$

Yes, A and B are independent.

Yes, they are independent.

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