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put this solution on YOUR website!A certain airplane has two independent alternators to provide electrical power. The probability that a given alternator will fail on a 1-hour flight is .02. What is that
(a) P(both will fail) = 0.02^2 0.0004
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(b) P(Neither will fail) = 0.08^2 = 0.0064
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(c) P(One or the other will fail) = 1-(0.0004+0.0064) = 0.9932
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Cheers,
Stan H.
You can
put this solution on YOUR website!P(A)=.02 P(B)=.02 A=Alternator #1 B=Alternator #2
a)P(A ∩ B) = P(A)P(B) = (.02)(.02) = .0004
b)(.98)(.98) = 0.9604
c)[P(alt 1 fails) * P(alt 2 doesn't fail)] + [P(alt 1 doesn't fail) * P(alt 2 fails)] = [ .02 * .98] + [.98 * .02] = .0392