Conditional Probability and Independent Events | NIMCET

Quick answer: The conditional probability of A given B is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$, for $P(B) > 0$. Rearranged, this is the multiplication theorem: $P(A \cap B) = P(B)\,P(A \mid B) = P(A)\,P(B \mid A)$. Events A and B are independent when $P(A \cap B) = P(A)\,P(B)$, which means knowing B does not change the probability of A.

This builds on the addition theorem of probability.

Conditional probability

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

Think of B as the new sample space. With equally likely outcomes, $P(A \mid B) = \frac{n(A \cap B)}{n(B)}$.

Multiplication theorem

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

For three events:

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

This is the rule behind drawing without replacement.

Independent events

Independent versus mutually exclusive: if $P(A) > 0$ and $P(B) > 0$, mutually exclusive events cannot be independent, because $P(A \cap B) = 0 \ne P(A)\,P(B)$.

With and without replacement

Worked examples

Example 1: from a table of outcomes

A family has two children. Given that at least one is a boy, find the probability that both are boys.

Step 1: Sample space BB, BG, GB, GG. The condition leaves BB, BG, GB.

Step 2: $P = \frac{1}{3}$.

Example 2: without replacement

A bag has 5 red and 3 black balls. Two are drawn without replacement. Find the probability that both are red.

$\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$

Check with combinations: $\frac{^5C_2}{^8C_2} = \frac{10}{28} = \frac{5}{14}$ ✓.

Example 3: independent events

A and B hit a target with probabilities $\frac{1}{2}$ and $\frac{1}{3}$, independently. Find the probability that the target is hit.

$1 - \frac{1}{2} \cdot \frac{2}{3} = 1 - \frac{1}{3} = \frac{2}{3}$

Example 4: testing independence

A die is rolled. A = "even", B = "multiple of 3". Are A and B independent?

$P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, $P(A \cap B) = P(\{6\}) = \frac{1}{6} = \frac{1}{2} \cdot \frac{1}{3}$. Yes.

Example 5: conditional from given values

If $P(A) = 0.4$, $P(B) = 0.5$ and $P(A \cap B) = 0.2$, find $P(A \mid B)$ and $P(B \mid A')$.

Step 1: $P(A \mid B) = \frac{0.2}{0.5} = 0.4$.

Step 2: $P(B \cap A') = 0.5 - 0.2 = 0.3$, so $P(B \mid A') = \frac{0.3}{0.6} = 0.5$.

(Here A and B are independent, since $0.4 \times 0.5 = 0.2$.)

Summary table

Exam tips

• Rewrite "given that" as the new sample space.
• For "at least one" with independent events, multiply the failure probabilities and subtract from 1.
• Check independence with the product rule, not intuition.

Next, read Bayes' theorem and the law of total probability, or see the NIMCET Probability and Statistics chapter.

Common mistakes

• Confusing $P(A \mid B)$ with $P(B \mid A)$.
• Treating draws without replacement as independent.
• Assuming mutually exclusive events are independent.
• Dividing by $P(A)$ instead of $P(B)$ in $P(A \mid B)$.

Frequently Asked Questions

What is the formula for conditional probability?

P(A | B) = P(A ∩ B)/P(B), provided P(B) > 0.

What is the multiplication theorem of probability?

P(A ∩ B) = P(A) · P(B | A).

When are two events independent?

When P(A ∩ B) = P(A) · P(B), or equivalently P(A | B) = P(A).

Are mutually exclusive events independent?

Not if both have positive probability. If one occurs, the other cannot, so the product rule fails.

What is the probability that at least one of two independent events occurs?

1 − P(A′)P(B′).