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Unformatted text preview: PROBABILITY AND STATISTICS, FALL 2008 SOLUTIONS FOR MIDTERM 1 Problem 1 Each of three closed boxes contains three coins. One box contains 3 gold coins, one contains three silver coins, and the third contains 1 gold and 2 silver coins. Suppose you choose a box purely at random and then choose a coin from that box at random. a) [5 pts] What is the probability that the chosen coin is gold? b) [5 pts] If the chosen coin really is gold, what is the probability that a second coin taken from the same box (without replacement of the first) will also be gold? Solution: a) Let Box 1=3G; Box 2=3S; Box3=1G2S. Define A = the chosen coin is gold. P ( A ) = P ( A  box 1) × P (Box 1) + P ( A  box 2) × P (Box 2) + P ( A  Box 3) × P (Box 3) ⇒ P ( A ) = 1 × 1 3 + 0 × 1 3 + 1 3 × 1 3 = 4 9 b) The key here is to recognize that the only way the second coin can be gold is when Box 1 is chosen. P (the second coin is gold  the first is gold) = P (Box 1  the first is gold) = P (the first is gold  Box 1) × P (Box 1) P (the first is gold) = 1 × 1 3 4 9 = 3 4 1 Problem 2 The random variables X and Y have the joint density f XY ( x,y ) = 6 xy for 0 ≤ x ≤ 1 and ≤ y ≤ √ x , and f XY ( x,y ) = 0 otherwise....
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 Spring '08
 Hayashi
 Variance, Probability distribution, Probability theory, probability density function

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