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Mathematics · lesson 1 of 2

Conditional probability and expectation

P(A∣B)=P(A and B)/P(B) requires P(B)>0.

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Learn the ideas

Prerequisites: Fractions, sums and elementary probability.

Conditioning

P(A∣B)=P(A and B)/P(B) requires P(B)>0.

Independence

A and B are independent when P(A and B)=P(A)P(B); independence is not the same as disjointness.

Expectation

Expectations add even without independence, provided the relevant expectations exist.

Variance

Variance adds for independent variables; in general include covariance terms.

Worked examples

Example 1

Given P(A and B)=0.12 and P(B)=0.4, find P(A∣B).

Divide 0.12 by 0.4.

0.3

Example 2

X takes values 0,2,5 with probabilities 0.2,0.5,0.3. Find E[X].

Compute 0(0.2)+2(0.5)+5(0.3)=2.5.

2.5

Check your understanding

8 questions

Decimals or fractions are accepted for numeric answers. Proofs and written responses use sample answers and require human marking. This is practice, not a protected examination.

1 Given P(A and B)=0.12 and P(B)=0.4, find P(A∣B).

Show answer and explanation

0.3

Divide 0.12 by 0.4.

2 X takes values 0,2,5 with probabilities 0.2,0.5,0.3. Find E[X].

Show answer and explanation

2.5

Compute 0(0.2)+2(0.5)+5(0.3)=2.5.

3 Two events have positive probability and are disjoint. Are they independent?

Show answer and explanation

No

Their intersection probability is zero while the product of their positive probabilities is positive.

4 Independent X and Y have variances 4 and 9. Find Var(2X−Y).

Show answer and explanation

25

Variance scales by the square of coefficients, giving 4×4+9=25.

5 Show by an example that E[XY] need not equal E[X]E[Y].

Sample answer and marking guidance

Let X=Y be a fair 0/1 variable. Then E[XY]=E[X]=1/2, while E[X]E[Y]=1/4.

Credit a dependent example and both computed quantities.

6 Does E[X+Y]=E[X]+E[Y] require independence?

Show answer and explanation

No, assuming expectations exist

Linearity of expectation does not require independence.

7 X is binomial with n=10 and p=0.3. Find its variance.

Show answer and explanation

2.1

np(1−p)=10(0.3)(0.7).

8 Explain why reversing a conditional probability without adjusting for base rates is generally wrong.

Sample answer and marking guidance

P(A∣B) and P(B∣A) have different conditioning populations. Bayes rule includes P(A) and P(B), so the two need not be equal.

Credit the change in denominator or conditioning population and the role of base rates.

Further reading and review references

These are reference links, not copied exercises, institutional endorsement or proof of independent review.