Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

no text concepts found

Transcript

PROBABILITY FLOW CHART EXPERIMENT S = SAMPLE SPACE (n(S) =TOTAL OUTCOMES IN SAMPLE SPACE ) A = EVENT (SUBSET OF S) (n(A) = TOTAL OUTCOMES IN EVENT) SIMPLE EVENTS (& COMPLEMENTARY EVENTS) MUTUALLY EXCLUSIVE NON-MUTUALLY EXCLUSIVE COMPLEX EVENTS INDEPENDENT CONDITIONAL Examples: Simple Event(& its complement): A box contains 2 black balls and 2 red balls. A ball is selected at random, its color recorded, and then it is replaced. A second ball is then selected at random, and its color recorded. Outcomes can be tabulated as shown below: R1 R2 B1 B2 B1 B1R1 B1R2 B1B1 B1B2 B2 B2R1 B2R2 B2B1 B2B2 R1 R1R1 R1R2 R1B1 R1B2 R2 R2R1 R2R2 R2B1 R2B2 The probability that both balls are black is 4/16 = ¼. The probability that both balls are not black = 1 – ¼ = ¾. Complex Events Mutually Exclusive: A single card is drawn at random from a standard deck of 52 cards. Let A = the card is a spade and B = the card is a heart A and B are mutually exclusive. P(A) = ¼ and P(B) = ¼. P(A or B) = P(A) + P(B) = ¼ + ¼ = ½ . Non-mutually exclusive: Two fair dice are rolled. Let A = the first die shows a two and B = the sum is 6 or 7 A and B are non-mutually exclusive. P(A) = 6/36 and P(B) = 11/36. P(A or B) = P(A) + P(B) – P( A and B ) = 6/36 + 11/36 – 2/36 = 15/36. Conditional: There are 25 fish in a pond. We know that 14 of these fish are males, 5 of these males are salmon, and there are 8 salmon in the pond. What is the probability that a randomly chosen fish is a salmon, given that it is a male? Solution: The condition ‘given that it is a male’ changes the size of the sample space under investigation. There are 14 males in all, so the size of the relevant sample space is 14. We also know 5 of these 14 are salmon. Hence, the probability that a randomly chosen fish is a salmon, given that it is a male is 5/14 Independent: When the outcome of one event has no influence on the outcome of a second event , the two events are said to be independent. a) A red die and a green die are rolled. Event A: red die shows odd Event B: green die show even. P(A and B) = P(A) * P(B) = ½ * ½ = ¼ b) Box I has two red balls and one black ball. Box II has one red ball and two black balls. Event A: red ball is drawn from Box I Event B: red ball is drawn from Box II P(A and B) = P(A) * P(B) = 2/3 * 1/3 = 2/9

Related documents