2.15 - ∪ S = P(F P(S – P(F)P(S = 0.7 0.85 – 0.7x0.85...

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2.15 F = fuel cell technology successfully marketable S = solar power technology successfully marketable F and S are statistically independent Given P(F) = 0.7; P(S) = 0.85 (i) P(energy supplied) = P(F
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Unformatted text preview: ∪ S) = P(F) + P(S) – P(F)P(S) = 0.7 + 0.85 – 0.7x0.85 = 0.955 (ii) P(only one source of energy available) = P(F S ∪ F S) = P(F S ) + P( F S) = (0.7)(1-0.85) + (1-0.7)(0.85) = 0.36...
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This note was uploaded on 04/25/2010 for the course MAE 213 taught by Professor Seshadri,k during the Spring '08 term at UCSD.

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