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Conditional Probability

Conditional Probability. The probability an event B will occur, given (on the condition) that another event A has occurred. We write this as P(B|A) and say “probability of B, given A.”.

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Conditional Probability

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  1. Conditional Probability The probability an event B will occur, given (on the condition) that another event A has occurred. We write this as P(B|A) and say “probability of B, given A.” Two cars are selected from a production line of 12 cars where 5 are defective. What is the probability the 2nd car is defective, given the first car was defective? Given a defective car has been selected, the conditional sample space has 4 defective out of 11. P(B|A) = 4/11

  2. Independent Events Two dice are rolled. Find the probability the second die is a 4 given the first was a 4. Original sample space: {1, 2, 3, 4, 5, 6} Given the first die was a 4, the conditional sample space is:{1, 2, 3, 4, 5, 6} The conditional probability, P(B|A) = 1/6

  3. Independent Events Two events A and B are independent if the probability of the occurrence of event B is not affected by the occurrence (or non-occurrence) of event A. A = 1st child is a boy B = 2nd child is a boy A = Being female B = Having type O blood Two events that are not independent are dependent. A = taking an aspirin each day B = having a heart attack A = being a female B = being under 64” tall

  4. Independent Events If events A and B are independent, then P(B|A) = P(B) Conditional Probability Probability 12 cars are on a production line where 5 are defective and 2 cars are selected at random. A = first car is defective B = second car is defective. The probability of getting a defective car for the second car depends on whether the first was defective. The events are dependent. Two dice are rolled. A = first is a 4 and B = second is a 4 P(B) = 1/6 and P(B|A) = 1/6. The events are independent.

  5. Contingency Table The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is: Omaha Seattle Miami Total Yes 100 150 150 400 No 125 130 95 350 Undecided 75 170 5 250 Total 300 450 250 1000 One of the responses is selected at random. Find: 1. P(Yes) 2. P(Seattle) 3. P(Miami) 4. P(No, given Miami)

  6. Solutions 300 450 250 Omaha Seattle Miami Total Yes 100 150 150 400 No 125 130 95 350 Undecided 75 170 5 250 Total 1000 = 400 / 1000 = 0.4 1. P(Yes) 2. P(Seattle) 3. P(Miami) 4. P(No, given Miami) = 450 / 1000 = 0.45 = 250 / 1000 = 0.25 = 95 / 250 = 0.38 Answers: 1) 0.4 2) 0.45 3) 0.25 4) 0.38

  7. Multiplication Rule To find the probability that two events, A and B will occur in sequence, multiply the probability A occurs by the conditional probability B occurs, given A has occurred. P(A and B) = P(A) x P(B|A) Two cars are selected from a production line of 12 where 5 are defective. Find the probability both cars are defective. A = first car is defective B = second car is defective. P(A) = 5/12 P(B|A) = 4/11 P(A and B) = 5/12 x 4/11 = 5/33 = 0.1515

  8. Multiplication Rule Two dice are rolled. Find the probability both are 4’s. A = first die is a 4 and B = second die is a 4. P(A) = 1/6 P(B|A) = 1/6 P(A and B) = 1/6 x 1/6 = 1/36 = 0.028 When two events A and B are independent, then P (A and B) = P(A) x P(B) Note for independent events P(B) and P(B|A) are the same.

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