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20065ee131A_1_EE131AF06HW1Solution

# 20065ee131A_1_EE131AF06HW1Solution - Solution of Assignment...

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Solution of Assignment 1, EE 131A Due: Wednesday October 11, 2006 Problem 1. (a) S = { ( F, F ) , ( F, R ) , ( F, K ) , ( R, F ) , ( R, R ) , ( R, K ) , ( K, F ) , ( K, R ) , ( K, K ) } . (b) A = { ( F, F ) , ( F, R ) , (( R, F ) , ( R, R ) } . Problem 2. (a) ( A B c C c ) uniontext( B A c C c ) uniontext( C A c B c ) . (b) ( A B C c ) uniontext( A C B c ) uniontext( B C A c ) . (c) A B C . (d) ( A B ) uniontext( A C ) uniontext( B C ) . (e) A c B c C c . Problem 3. We use the identity P [ A B ] = P [ A ] + P [ B ] - P [ A B ] , for the sets A = { a, c } and B = { b, c } . The result is 1 = P [ { a, b, c } ] = 5 8 + 7 8 - P [ c ] . Therefore, P [ c ] = 1 2 . Since, P [ { a, c } ] = P [ a ] + P [ c ] and P [ { b, c } ] = P [ b ] + P [ c ] , thus P [ a ] = 1 8 and P [ b ] = 3 8 . Problem 4. (a) We use the inequality P [ X Y ] P [ X ] + P [ Y ] , then P bracketleftbig A B C bracketrightbig = P bracketleftbig ( A B ) C bracketrightbig P [ A B ] + P [ C ] P [ A ] + P [ B ] + P [ C ] . (b) If A B = then A B c , therefore P [ A ] P [ B c ] . Problem 5. 8 × 10 × 10 × 10 × 10 × 10 × 10 = 8 × 10 6 . Problem 6. In the case that toppings can be repeated this is the same as sampling with replace- ment and without order which is equal to ( 15 - 1+4 4 ) = 3060 . In the case that repetitions is not possible, there are ( 15 4 ) = 1365 combinations.

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Problem 7. The number of arrangements of 20 cards in a row is the same as the number of
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20065ee131A_1_EE131AF06HW1Solution - Solution of Assignment...

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