G 3 points is the number of sixes in twenty rolls

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(g) [3 points] Is the number of sixes in twenty rolls independent from the number of ones? Explain. Solution: They are not independent. For example P (19 sixes and 2 ones) = 0 , but P (19 sixes) P (2 ones) 6 = 0. 2. There are two boxes. The first contains 5 red balls and 1 black ball. The second contains 2 red balls and 1 black ball. You pick a box and then pick two balls from that box. (a) [5 points] What is the probability that both are red? Solution: 1 2 × 5 6 4 5 + 1 2 × 2 3 1 2 = 1 2 . (b) [5 points] What is the probability that the second is red given that the first is red? Solution: P (second red | first red) = P (both red) P (first red) = 1 2 1 2 × 5 6 + 1 2 × 2 3 = 2 3 . (c) [5 points] What is the probability that you picked the first box given that both balls are red? Solution: P (first box | both red) = P (first box and both red) P (both red) = 1 2 × 5 6 4 5 1 2 = 2 3 . Page 2
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Math 431: Exam 1 3. Pick 5 cards out of a standard deck of 52 cards. (a) [5 points] What is the chance that there are at least two hearts? Solution: 1 - P (no hearts) - P (one heart) = 1 - ( 39 5 ) ( 52 5 ) - ( 13 1 )( 39 4 ) ( 52 5 ) 0 . 367 . (b) [5 points] In how many suits do you expect to have more than one card? Solution: Let I H , I C , I D and I S be indicators, for each suit, that there is more than one card in that suit. The number of suits with more than one card is X = I H + I C + I D + I S . Note that we calculated E ( I H ) = P (more than one heart) above. The expectations of I C , I D and I S are the same. Hence E ( X ) = 4 E ( I H ) = 4 1 - ( 39 5 ) ( 52 5 ) - ( 13 1 )( 39 4 ) ( 52 5 ) ! 1 . 468 . 4. [8 points] A hat contains two black and two white cards. Pick two cards and let X be the number of black cards. Find E ( X ) and SD ( X ).
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