Multiplication Theorem of Probability

Define events in order. Decide whether the trial is with or without replacement. Write the probability of the first event. Write the probability of the next event under the previous condition. Multiply the branch probabilities and simplify.

A tree diagram is useful for successive trials or staged choices. Each branch should show the conditional probability after the previous outcome.

Using (4/10)(4/10) in a without-replacement problem, ignoring that the total and favorable counts change after the first selection.

For two events A and B, P(A∩B) = P(A)P(B|A) when P(A) > 0, and also P(A∩B) = P(B)P(A|B) when P(B) > 0.

For two events A and B, P(A∩B) = P(A)P(B|A) when P(A) > 0, and also P(A∩B) = P(B)P(A|B) when P(B) > 0. Key result: P(A∩B)=P(A)P(B|A)=P(B)P(A|B). For three events, P(A∩B∩C)=P(A)P(B|A)P(C|A∩B), with required conditioning probabilities defined. A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Probability that both are red = P(first red)P(second red | first red) = (5/8)(4/7) = 5/14.

The multiplication theorem rewrites the probability of both events occurring together using one event first and the other event under that condition. It is especially useful in word problems where probabilities are given in stages. Key result: P(A∩B)=P(A)P(B|A)=P(B)P(A|B). For three events, P(A∩B∩C)=P(A)P(B|A)P(C|A∩B), with required conditioning probabilities defined. A box contains 4 defective and 6 good items. Two items are selected without replacement. Find the probability both are defective. Let A be first defective and B be second defective. P(A)=4/10. After A, 3 defective remain out of 9, so P(B|A)=3/9. Thus P(A∩B)=(4/10)(3/9)=12/90=2/15. It is commonly asked in drawing balls, selecting cards, quality control, family or committee problems, and multi-stage probability situations. Using (4/10)(4/10) in a without-replacement problem, ignoring that the total and favorable counts change after the first selection.