Simulation · Maths · Class 12

Shrink the sample space

From the lesson Conditional probability in Probability. Change the values and watch what happens.

The idea behind it

NCERT §13.1–13.2, Examples 1–3

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.

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