3.4 Conditional Probability and Independent Events

Given information changes the sample space

Conditional probability asks for the probability of one event after another event is already known. The notation P(B|A) is read as probability of B given A. The given event A becomes the new sample space. This is why a two-way table is often easier than a formula: cover everything outside the given row, column, or group, then count the part that also satisfies the target event.

The formula is P(B|A) = P(A and B) / P(A), as long as P(A) is not zero. In count form, it is even more direct: count outcomes that are both A and B, then divide by the count of outcomes in A.

Worked example: two-way table

A school survey records transportation for 120 students. Of 70 Grade 12 students, 28 drive and 42 do not drive. Of 50 Grade 11 students, 12 drive and 38 do not drive. What is the probability that a student drives, given that the student is in Grade 12?

The phrase given that the student is in Grade 12 restricts the denominator to 70 Grade 12 students. Within that group, 28 drive. The probability is 28/70 = 2/5 = 0.40.

Do not use 40/120, even though 40 students drive overall. That answers the probability a randomly selected student from the whole survey drives. The condition has already told you the student is in Grade 12, so the whole-school denominator is no longer correct.

Independent events

Two events are independent if knowing one occurred does not change the probability of the other. Algebraically, P(A and B) = P(A)P(B). Conceptually, P(B|A) = P(B).

A common example is drawing with replacement from the same bag. If a marble is replaced and the bag is mixed before the next draw, the composition returns to the original state. The second probability is the same as the first. Without replacement, the composition changes, so the events are usually dependent.

Worked example: A bag has 5 red and 7 blue counters. Two counters are drawn without replacement. What is the probability both are red?

First red: 5/12. After one red is removed, 4 red counters remain out of 11 total counters. Second red given first red: 4/11. Multiply along the branch: (5/12)(4/11) = 20/132 = 5/33.

If the first counter were replaced, the probability would be (5/12)(5/12), because the second draw would again have 5 red counters out of 12. The phrase without replacement is small, but it changes the entire setup.

Testing independence from data

Sometimes a table gives data and asks whether two events are independent. Compare the overall probability of an event with the conditional probability inside the given group.

Suppose 30 of 100 students take drama, and among 40 Grade 12 students, 12 take drama. Overall P(drama) = 30/100 = 0.30. Conditional P(drama | Grade 12) = 12/40 = 0.30. Because the probabilities match, taking drama and being in Grade 12 are independent in this survey data.

If the conditional probability were 18/40 = 0.45, the events would not be independent because knowing the student is in Grade 12 changes the probability of drama participation.

At least one with independent trials

Repeated independent trials often pair with the complement rule. If the probability that a student solves a puzzle on one try is 0.65 and tries are treated as independent, the probability the student solves it at least once in three tries is 1 - P(no success in all three tries). The probability of no success on one try is 0.35, so the result is 1 - (0.35)^3 = 0.957125.

Listing one success, two successes, and three successes is possible, but it is longer and easier to miscount. Complement thinking is usually faster for at least one.

Diploma traps

Trap 1: using the original denominator after a condition. Given means the sample space has changed. Circle or underline the given group before calculating.

Trap 2: assuming without replacement is independent. If an item is removed, update both the favourable count and the total count before the next probability.

Trap 3: confusing independent with mutually exclusive. Mutually exclusive events cannot happen together. Independent events can happen together, but one does not change the chance of the other. Two non-impossible mutually exclusive events are not independent because if one happens, the other has probability zero.

Trap 4: multiplying probabilities before checking the condition. Multiplication is correct for an and path only after each branch probability has the right denominator. For dependent events, the second factor is conditional.

A strong Math 30-2 solution writes the branch or table denominator beside every probability. That habit prevents the most common errors with conditional probability and dependent events.