lecture04-sep17

# lecture04-sep17 - STAT 430 Probability Lecture 4 Fall 2007...

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STAT 430 Probability Lecture 4 Fall, 2007

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Homework Due in Class on Monday, September 24: Section 1.4: 4, 6, 7* Section 1.5: 2, 4, 6. Section 1.6: 2, 3*, 4*, 6 Problems with * are not graded.
Example A family has two children. Assume all four possible “outcomes” -- (younger is a boy, older is a girl), …, are equally likely. What is a probability that both are boys given that at least one is a boy? Let A = Both children are boys, B = At least one is a boy. Then P( A ) = 1/4, P( B ) = 3/4. (Why?) P(B c )=P(both are girls) = 1/4. P(B)= 1- P(B c ) = 3/4. So, P( A | B )=P( A B )/P( B ) = P( A )/P( B ) = 1/3.

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An urn contains 6 white balls and 5 blue chips. Two balls are drawn sequentially without replacement. What is the probability of obtaining the sequence (white, blue)? A = 1 st ball is white; B = 2 nd ball is blue. Want to know P(A B) = ? Use the formula P(A B) = P(B | A) P(A) P(A) = 6/11. P(B | A) = 5/10 = 1/2. So,
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lecture04-sep17 - STAT 430 Probability Lecture 4 Fall 2007...

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