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A calculator requires a keystroke assembly and a logic circuit. Assume that 83% of the keystroke assemblies and 88% of the logic circuits are satisfactory. Assuming that defects in keystroke assemblies are independent of defects in logic circuits, find the probability that a finished calculator will be satisfactory. Group of answer choices

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Answer:

0.7304

Step-by-step explanation:

According to the Question,

  • Given That,  A calculator requires a keystroke assembly and a logic circuit. Assume that 83% of the keystroke assemblies and 88% of the logic circuits are satisfactory.

We have,

P(keystroke satisfactory) =0.83 ,  P(logic satisfactory)= 0.88

  • Assuming that defects in keystroke assemblies are independent of defects in logic circuits

Since the events are independent. So, the probability that a finished calculator will be satisfactory

⇒ 0.83×0.88

⇒0.7304

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