Lesson 1 of 10 · 11 min

Conditional probability

NCERT §13.1–13.2, Examples 1–3

Asha's bus is late on 32 mornings out of 100. She looks out of the window and sees rain. Is 0.32 still the right number?

The story this chapter follows: Asha's school bus

Asha takes the school bus every morning and has kept a record of 100 mornings: whether it rained and whether the bus was late. The chapter follows her as she asks what rain tells her about the bus, and what a late bus tells her about the rain. All numbers are made up for the story.

The lesson in notes

In short

When it is known that an event F has occurred, the outcomes outside F are no longer possible. F becomes the new sample space, and the probability of E is recomputed inside it. The result is the conditional probability of E given F, written P(E|F).

Three fair coins are tossed: 8 equally likely outcomes. E: at least two heads = {HHH, HHT, HTH, THH}; F: the first coin shows tail = {THH, THT, TTH, TTT}. P(E) = P(F) = 1/2 and E ∩ F = {THH}, so P(E ∩ F) = 1/8.

Given F, only its four outcomes remain, and just one of them (THH) lies in E. So P(E|F) = 1/4, not 1/2: the information has changed the probability.

With equally likely outcomes, P(E|F) = n(E ∩ F)/n(F). Dividing top and bottom by n(S) gives the form used in general.

Definition: P(E|F) = P(E ∩ F) ÷ P(F), which needs P(F) ≠ 0.

With P(A ∩ B) = 4/13 and P(B) = 9/13, the definition gives P(A|B) = (4/13) ÷ (9/13) = 4/9. The value P(A) = 7/13 given alongside is not needed for this.

A family has two children, so S = {bb, bg, gb, gg}. Given that at least one is a boy (3 outcomes), the probability that both are boys is 1/3.

Cards numbered 1 to 10; one is drawn and is known to be more than 3, leaving {4, 5, …, 10}. Four of these seven are even, so P(even | more than 3) = 4/7.

P(E|F) and P(F|E) are different quantities: they share a numerator but are divided by P(F) and P(E) respectively.