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# solution02 - Solutions to Homework 02 1 We are interested...

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Solutions to Homework 02 1. We are interested in the shaded region shown in Figure 1. The shaded are on the left corresponds to A B C while that on the right is A C B . Also, the unshaded area common to both A and B is A B . Now, note that A = ( A B C ) ( A B ) P [ A ] = P [ A B C ] + P [ A B ] P [ A B C ] = P [ A ] - P [ A B ] similarly P [ A C B ] = P [ B ] - P [ A B ] Then, P [( A B C ) ( A C B )] = P [ A B C ] + P [ A C B ] = P [ A ] - P [ A B ] + P [ B ] - P [ A B ] = P [ A ] + P [ B ] - 2 P [ A B ] A B Figure 1: Venn diagram for question 1. 2. Refer to Figure 2(a) for the region A C B C . P [ A C B C ] = 1 - P [ A B ] = 1 - z 1

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Refer to Figure 2(b) for the region A B C . P [ A B C ] = P [ A ] - P [ A B ] = x - z Refer to Figure 2(c) for the region A C B . P [ A C B ] = 1 - P [ A B C ] = 1 - ( x - z ) = 1 - x + z Refer to Figure 2(d) for the region A C B C . P [ A C B C ] = 1 - P [ A B ] = 1 - ( P [ A ] + P [ B ] - P [ A B ]) = 1 - x - y + z (a) (b) (c) (d) Figure 2: Venn diagram for question 1. 3. Let x denote the lifetime. Then A = { x : x > 5 } B = { x : x > 10 } (a) P [ A B ] = [ { x > 5 } ∩ { x > 10 } ] = P [ { x > 10 } ] = e - 10 2
P [ A B ] = [ { x > 5 } ∪ { x > 10 } ] =

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solution02 - Solutions to Homework 02 1 We are interested...

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