document.write( "Question 161724: A calculator requires a keystroke assembly and a logic circuit. Assume that 88% of the keystroke assemblies and 93% of the logic circuits are satisfactory.\r
\n" ); document.write( "\n" ); document.write( "a. Find the probability that a finished calculator will be satisfactory\r
\n" ); document.write( "\n" ); document.write( "b. Find the probability that a finished calculator will have both a bad keystroke assembly and a bad logic circuit.\r
\n" ); document.write( "\n" ); document.write( "c. Find the probability that a finished calculator will have either a bad keystroke assembly or a bad logic circuit.
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Algebra.Com's Answer #119197 by gonzo(654)\"\" \"About 
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let a = key stroke assembly is good.
\n" ); document.write( "let b = logic circuit is good
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\n" ); document.write( "p(a) = .88
\n" ); document.write( "p(~a) = 1-.88 = .12
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\n" ); document.write( "p(b) = .93
\n" ); document.write( "p(~b) = .07
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\n" ); document.write( "~a means (not) a means keystroke assembly is bad
\n" ); document.write( "~b means (not) b means logic circuit is bad.
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\n" ); document.write( "a. Find the probability that a finished calculator will be satisfactory
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\n" ); document.write( "probability finished calculator is good means the keystroke assembly and the logic circuit is good.
\n" ); document.write( "p(a and b) = p(a)*p(b)
\n" ); document.write( "p(a)*p(b) = .88*.93 = .8184
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\n" ); document.write( "b. Find the probability that a finished calculator will have both a bad keystroke assembly and a bad logic circuit.
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\n" ); document.write( "probability that a finished calculator will have both a bad keystroke assembly and a bad logic circuit is p (~a and ~b).
\n" ); document.write( "p(~a and ~b) = p(~a)*p(~b) = .12*.07 = .0084
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\n" ); document.write( "c. Find the probability that a finished calculator will have either a bad keystroke assembly or a bad logic circuit.
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\n" ); document.write( "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.
\n" ); document.write( "p(~(a or b)) = 1 - p(a and b)
\n" ); document.write( "p(a and b) = p(a)*p(b) = .8184 (calculated above in part a).
\n" ); document.write( "1 - p(a and b) = 1 - p(a)*p(b) = 1 - .8184 = .1816.
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