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solutions02

# solutions02 - Math 302.102 Fall 2010 Solutions to...

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Math 302.102 Fall 2010 Solutions to Assignment #2 1. Let R j , j = 1 , 2 , . . . , 50, be the event that the j th box of Raisin Bran contains 2 scoops of raisins. We are told that P { R j } = 0 . 93 for each j and that the events R 1 , R 2 , . . . , R 50 are independent. (a) The probability that all of Danny’s 50 packages contain two scoops of raisins is P { R 1 R 2 ∩ · · · ∩ R 50 } = P { R 1 } P { R 2 } · · · P { R 50 } = (0 . 93) 50 . (b) Notice that { at least one of Danny’s 50 packages does not contain two scoops } c = { none of Danny’s 50 packages do not contain two scoops } = { all of Danny’s 50 packages contain two scoops } This implies that P { at least one of Danny’s 50 packages does not contain two scoops } = 1 - P { all of Danny’s 50 packages contain two scoops } = 1 - (0 . 93) 50 . Symbolically, the event { at least one of Danny’s 50 packages does not contain two scoops } can be written as R c 1 R c 2 ∪ · · · ∪ R c 50 and so P { R c 1 R c 2 ∪ · · · ∪ R c 50 } = 1 - P { ( R c 1 R c 2 ∪ · · · ∪ R c 50 ) c } = 1 - P { R 1 R 2 ∩ · · · ∩ R 50 } = 1 - (0 . 93) 50 using the fact that ( A c B c ) c = ( A c ) c ( B c ) c = A B . (This is sometimes called De Morgan’s law.) (c) Notice that there are several ways in which exactly one package that does not contain two scoops of raisins could happen. It might be that only the first box does not contain two scoops, or it might be that only the second box does not contain two scoops, etc. Thus, P { exactly one of Danny’s 50 packages does not contain two scoops } = P { ( R c 1 R 2 ∩ · · · ∩ R 50 ) ( R 1 R c 2 R 3 ∩ · · · ∩ R 50 ) ∪ · · · ∪ ( R 1 R 2 ∩ · · · ∩ R c 50 ) } = P { R c 1 R 2 ∩ · · · ∩ R 50 } + P { R 1 R c 2 R 3 ∩ · · · ∩ R 50 } + · · · + P { R 1 R 2 ∩ · · · ∩ R c 50 } = (0 . 07)(0 . 93) · · · (0 . 93) + (0 . 93)(0 . 07)(0 . 93) · · · (0 . 93) + · · · + (0 . 93) · · · (0 . 93)(0 . 07) = (0 . 07)(0 . 93) 49 + (0 . 07)(0 . 93) 49 + · · · + (0 . 07)(0 . 93) 49 = 50(0 . 07)(0 . 93) 49 .

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2. (a) Let D j be the event that the j th song is by Snoop Dogg. We are interested in computing P { D 1 D 2 D 3 } . The events D 1 , D 2 , and D 3 are NOT independent since we are not allowed to repeat a song in the playlist. Intuitively, there is a 3/20 chance that the first song is by Snoop Dogg. Given that the first song is by Snoop Dogg, there is a 2/19 chance that the second song will
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