Question

A calculator requires a keystroke assembly and a logic circuit. Assume that 80% of the keystroke assemblies and 84% of the logic circuits are satisfactory. Find the probability that a finished calculator will be satisfactory. Assume that defects in keystroke assemblies are independent of defects in logic circuits. O A. 0.7056 OB. 0.6720 OC. 0.6400 OD. 0.8200

Answer 1

When two events are independent (say A & B ),
P( A&B ) = P(A) * P(B)

Thus,
P(Finished calculator is satisfactory ) = P( Keystroke assembly is satisfactory ) * P( Logic circuits is satisfactory)
= 0.80 * 0.86
= 0.672

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