<PRE><B>Question 161724</B>
let a = key stroke assembly is good.
let b = logic circuit is good
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p(a) = .88
p(~a) = 1-.88 = .12
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p(b) = .93
p(~b) = .07
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~a means (not) a means keystroke assembly is bad
~b means (not) b means logic circuit is bad.
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a. Find the probability that a  finished calculator will be satisfactory
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probability finished calculator is good means the keystroke assembly and the logic circuit is good.
p(a and b) = p(a)*p(b)
p(a)*p(b) = .88*.93 = .8184
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b. Find the probability that a finished calculator will have both a bad keystroke assembly and a bad logic circuit.
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probability that a finished calculator will have both a bad keystroke assembly and a bad logic circuit is p (~a and ~b).
p(~a and ~b) = p(~a)*p(~b) = .12*.07 = .0084
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c. Find the probability that a finished calculator will have either a bad keystroke assembly or a bad logic circuit.
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probability that a finished calculator will have either a bad keystroke assembly or a bad logic circuit is equal to 1 minus the probability that they are both good.
p(~(a or b)) = 1 - p(a and b)
p(a and b) = p(a)*p(b) = .8184 (calculated above in part a).
1 - p(a and b) = 1 - p(a)*p(b) = 1 - .8184 = .1816.
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