Lesson 3 of 10 · 7 min
Multiplication theorem on probability
NCERT §13.3, Examples 8–9
What is the chance that tomorrow it rains and the bus is late? Two things must both happen.
The lesson in notes
In short
E ∩ F, also written EF, is the event that both E and F occur.
Clearing the fraction in the definition of conditional probability gives the multiplication rule: P(E ∩ F) = P(E) P(F|E), and equally P(F) P(E|F). The first form needs P(E) ≠ 0, the second P(F) ≠ 0.
Read it as a sequence: the chance that E happens, times the chance that F then happens once E has.
An urn holds 10 black and 5 white balls; two are drawn in turn without replacement. P(first black) = 10/15. With one black gone, P(second black | first black) = 9/14. So P(both black) = (10/15)(9/14) = 3/7.
For three events: P(A ∩ B ∩ C) = P(A) P(B|A) P(C|A ∩ B). The rule extends the same way to four or more events.
Three cards are drawn in turn without replacement from a well-shuffled pack of 52. P(king, king, ace in that order) = (4/52)(3/51)(4/50) = 2/5525.
Without replacement, each draw changes the counts, so every factor after the first is a conditional probability. With replacement the counts are restored and the factors do not change.