Lecture 12 Sections 5.3

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## Lecture 12 Sections 5.3

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**Lecture 12Sections 5.3**Objectives: • Conditional Probability • Definition • Independent Events**Conditional Probability**Consider two events, A and B. Suppose that we know that B has occurred. This knowledge may change the probability that A will occur. We denoted by P(A|B) the conditional probability of event A given that B has occurred. Example. Suppose that a card is randomly drawn from a well-shuffled deck of 52 cards. Then, the probability of drawing a Heart is ¼. But, if additional information is given that the drawn card is red, then the probability of a Heart changes to ½, because the sample space is now reduced to just 26 red cards, of which 13 are Hearts. Thus, P(Heart|Red)=1/2. On the other hand, if we are told that the drawn card is black, then P(Heart|Black)=0.**Conditional Probability**Definition: The conditional probability of an event A given an event B with P(B) > 0 is defined as The knowledge that B has occurred effectively reduces the sample space from S to B. Therefore, P(A|B) is interpreted as the proportion of the area of B occupied by A. Note: For two events A and B with P(A)>0 and P(B)>0, P(A∩ B) = P(A | B)P(B) or P(A∩ B) = P(B | A)P(A) from the definition of the conditional probability. Example) Revisit the previous card example. Then, P(Heart|Red)=?**Example**• In a CD company two machines produce CD’s. • 60% of CD’s are produced by machine 1 (M1) • 40% of CD’s are produced by machine 2 (M2) • 10% of CD’s produced by M1 are defective • 5% of CD’s produced by M2 are defective • What is the probability that a CD is produced by M1 and it is defective? • b. What is the probability that a CD is defective? • c. What is the probability that a CD is produced by M2 given that it is defective?**Independent Events**• The probability that any two events, A and B, both occur is: P(A and B) = P(A)P(B|A) This is the general multiplication rule. • If A and B are independent, then P(A and B) = P(A)P(B) (A and B are independent when they have no influence on each other’s occurrence.) What is the probability of randomly drawing either an ace or heart from a deck of 52 playing cards? There are 4 aces in the pack and 13 hearts. P(heart|ace) = 1/4 P(ace) = 4/52 • P(ace and heart) = P(ace)* P(heart|ace) = (4/52)*(1/4) = 1/52 Notice that heart and ace are independent events.**Examples**Example) Suppose that the switches A and B in a two-component series system are closed about 60% and 80% of the time, respectively. If we assume that the closing of switch A occurs independently of switch B, find the probability that the entire circuit is closed. Example) Consider the following series and parallel circuit. Closed circuit is when flow reaches from Enter to Exit. Assume that switches function independently. Find the probability that the circuit is closed.