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Previous Lecture: We learnt the following: Random Experiment. Sample Space. An Event. The Probability. This lecture we will learn Conditional Probability. Independent Events. Multiplication Rules. Conditional Probability We often want to find the probability of an event A occurring given that B has occurred. In these cases, we are conditioning on B, and we write P(A|B). B serves as a new (reduced) sample space. P(A|B) is that fraction of P(B) which corresponding to P(A ∩ B). Definition: The conditional probability of B, given A, denoted by P(A|B) is defined by Where P(B) > 0. Example(conditional Probability): Throw two dice. The sum shown is 10. What is the probability that one 6 is cast? Solution: Let A be the event “ one 6 is cast”. Let B the event “the sum equals 10”. We want to calculate P(A|B). A={(1,6 ) , (2,6), (3,6 ) ,(4,6), (5,6) ,(6,6) , (6,1) , (6,2), (6,3) ,(6,4), ( 6,5)}. B= {(6,4), (4,6), (5,5)}. Example cont. P (A ∩ B) = 2/36. P(B) = 3/36. And we know So: P(A|B)= 2/3. Example If we have P(A)=0.35, P(B)=0.5, P(A⋃B)=0.8, Find P(A⋂B), P(A|B), P(B|A) P(A⋂B)=P(A)+P(B)-P(A⋃B) 0.35+0.5-0.8=0.05 P(A|B)= 𝑃(𝐴⋂𝐵) 𝑃(𝐵) 0.05/0.5 P(B|A)= 𝑃(𝐴⋂𝐵) 𝑃(𝐴) Example in a class of 30 students ,there are 17 girls and 13 boys .A total of 5 students made an A grade on their report card , three of which were girls .if a student is a chosen at random . The probability of getting a girl or an A student is …….. Independence A and B are independent if knowledge that one event has occurred does not change the probability that the other will occur. Definition: Two events A and B are independent if and only if P(A|B)= P(A) or P(B|A)=P(B) Provided the existences of the conditional probability. Otherwise, A and B are dependent. Example: Classifying Events as Independent or dependent: Tossing a coin and getting a head(A), and then rolling a die and obtaining a 6 (B). Driving over 85 miles per hour (A), and then getting in a car accident(B). Exercising frequently (A) and having a low blood pressure. Solutions: P(B|A)= 1/6 and P(B)= 1/6. The occurrence of A does not change the probability of occurrence of B, so the events are independent. If you drive over 85 miles per hour, the chances of getting in car accident are greatly increased, so these events are dependent. If you exercise regularly it will affect the blood pressure by make it lower, so these events are dependent. Example Let A and B are independent events as follows: P(A)= 0.5, and P(B)=0.6, Find P(A|B), P(B|A). Solution: Because the event are independent P(A|B)=P(A)= 0.5, and P(B|A)=P(B)=0.6. Multiplicative Rules Theorem: If in an experiment the events A and B can both occur, then : P (A ∩ B)= P(A)P(B|A) , provided that P(A) >0. Example: Two cards are selected, without replacing the first card from a standard deck. Find the probability of selecting a king and then selecting a queen. Solution: Because the first card not replaced, the events are dependent P (K ∩ Q)= P(K)P(Q|K) =4/52 . 4/51 =16/2652 ≈ 0.006. Theorem: Two events A and B are independent if and only if P (A ∩ B)= P(A)P(B) . Therefore, to obtain the probability that two independent events both occur, we simply find the product of their individual probabilities. Example: A coin is tossed and a die is rolled. Find the probability of getting a head and then rolling a 6. Solution: The events are independent. P(A∩ B)= P(A). P(B) =1/2.1/6=1/12≈ 0.083 So the probability of tossing a head and then rolling a 6 is about 0.083. Example A box contains 5 white balls, and 7 blue balls. If two balls were randomly chosen (one after another), calculate the following probabilities: -The two balls are white, chosen with replacement. -The two balls are blue, chosen without replacement. Example Let A and B be two events, with probabilities P(A) = 0.6, and P(B) =0.2 Calculate the following probabilities: P(A`), P(A⋂B), P(A⋃B), P(A|B), P(B|A), if: A and B are mutually exclusive (disjoint). • A and B are independent. • Example The probability that a salmon swims successfully through a dam is 0.85. Find the probability that three salmon swim successfully throw the dam. Solution The probability that each salmon is successful is 0.85, one salmon's chance of success is independent of the others chances P(3 salmon successful)= (0.85). (0.85).(0.85) ≈0.614 Example Find the probability that none of the three salmon is successful. Solution Because the probability of success for one salmon is 0.85, the probability of failure for one salmon is 1 -0.85=0.15 P(none of the three is successful) = (0.15).(0.15).(0.15) ≈0.003 So the probability of none are successful is about 0.003.