Question

Of the coffee makers sold in an appliance store, 4% have either a faulty switch or a defective cord, 1.3% have a faulty switch, and 0.4% have both defects. What is the probability that a coffee maker will have a defective cord given that it has either a faulty switch or defective cord? Give results to the nearest tenth of a percent.

Answer 1

Let P(A) be the probability of a faulty switch only. P(A) = 0.013
Let P(B) be the probability of a defective cord only. P(B) = 0.04 - 0.013 = 0.027
Let P(C) be the probability of both defects. P(C) = 0.004
The required probability is given by:
\[\frac{P(B)+P(C)}{P(A)+P(B)+P(C)}\times\frac{100}{1}=you\ can\ calculate\]