Conditional Probability and Independent Events: Quick Revision

A brief note on Conditional Probability and Independent Events. Read it before a test. Try each question at the end before you open its answer.

Conditional probability

• Definition: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$, with $P(B) > 0$. Indian books write $P(A/B)$.
• Count form: for equally likely outcomes, $P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}$.
• Complement: $P(A' \mid B) = 1 - P(A \mid B)$.
• Swapping: $P(A \mid B)$ and $P(B \mid A)$ have the same top and different bottoms.

Multiplication rule and trees

• Two events: $P(A \cap B) = P(A)\,P(B \mid A)$.
• Three events: $P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B)$.
• Tree: multiply along a path, add across paths, and the branches from a point add up to $1$.
• At least one: $1 - P(\text{none})$.

Independent events

• Test: $P(A \cap B) = P(A)\,P(B)$, or $P(A \mid B) = P(A)$.
• Complements: if $A$ and $B$ are independent, so are $A$ and $B'$, and $A'$ and $B'$.
• Not the same as exclusive: mutually exclusive events with positive probabilities are dependent.
• Three events: mutual independence needs the three pairs and the triple product.

A worked example

A bag holds $5$ red and $3$ green balls. Two are drawn without replacement.

1. Both red. $\dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{5}{14}$.
2. Second red, given first red. $\dfrac{4}{7}$, because $4$ red balls remain among $7$.
3. Second red, given first green. $\dfrac{5}{7}$, because $5$ red balls remain among $7$.

The two conditional answers differ, so the two draws are dependent.

Practice with new numbers

A program to try

The program finds the three answers of the worked example with exact fractions.

Mistakes to avoid

• Dividing by the wrong event. Divide by the event after the bar.
• Treating draws without replacement as independent. The contents change after the first draw.
• Adding for "and". Multiply along a path.
• Calling exclusive events independent. They are dependent when both have positive probability.

Check yourself

1. Find $P(A \mid B)$ when $P(A \cap B) = 0.12$ and $P(B) = 0.3$.

2. Two cards are drawn from a pack of $52$ without replacement. Find the probability that both are hearts.

3. Events $A$ and $B$ are independent, with $P(A) = 0.5$ and $P(B) = 0.4$. Find $P(A \cup B)$.

4. A fair coin is tossed $5$ times. Find the probability of at least one head.

5. Can two mutually exclusive events with positive probabilities be independent?

Sources

• OpenStax, Introductory Statistics, Chapter 3, sections on the rules of probability, contingency tables and tree diagrams.
• NCERT, Mathematics Part II for Class XII, Chapter 13, sections on conditional probability, the multiplication theorem and independent events.