Probability.pptVIP

General Multiplication Rule The Conditional Probability Rule states that for any two events A and B, P(B | A) = P(AB) P(A) If we rewrite this expression we have the General Multiplication Rule: P(AB) = P(A) · P(B | A) When A and B are independent events, then P(AB) = P(A) · P(B) If two cards are drawn consecutively without replacement, what is the probability that the first card will be a jack and the second card will be a face card? This is a two-part task with dependent events because the outcome of the first event (drawing the 1st card) affects the outcome of the second event (drawing the 2nd card). P(jack face card) = P(jack) · P(face | jack) = 4/52 · 11/51 = 44/2652 = 11/663 ? 0.01659 Multiplication Rule – Dependent Events If five cards are drawn consecutively without replacement, what is the probability that all five cards are face cards? This is a five-part task with dependent events because the outcome of the first event (drawing the 1st card) affects the outcome of the second event (drawing the 2nd card), which affects the outcome of the third event (drawing the 3rd card), etc. Thus P(5 face cards) is P(face) · P(face | face) · P(face | 2face) · P(face | 3face) · P(face | 4face) = 12/52 · 11/51 · 10/50 · 9/49 · 8/48 = 33/108,290 ? 0.0003047 Note that this can also be done as C(12,5) = 792 C(52,5) 2,598,960 Multiplication Rule – Dependent Events Multiplication Rule – Dependent Events An urn contains two blue marbles, four red marbles and five white marbles. If you draw four marbles without replacement, what is the probability that you select a red, two white and one blue marble in that order? This is a four-part task with dependent events since the sample space is reduced by one marble after each pick. Let’s define the events: R = pick red marble W = pick white marble B = pick blue marble P(red on 1st white on 2nd white on 3rd blue on 4th) = P(RWWB) = P(R)·P(W | R)·P(W | RW) · P(B | RWW) = 4