Question

6. Heart failures are due to either natural occurrence (87%) or outside factors (13%). Outside factors are related to induced substances (73%) or foreign objects (27%). Mat ural occurrences are caused by arterial blockage (56%), disease (27%), and infection (eg, staph infection) (17%). (a) Determine the probability that a failure is due to an induced substance. (b) Determine the probability that a failure is due to disease or infection.

Answer 1

6. (a) Supposing there are n occurrences of failure. Then, due to outside factors, there would be $\dpi{80} 0.13n$ occurrences. Out of those, there would be $\dpi{80} 0.13n*0.73$ occurrences due to induced substances. Hence, the probability that a failure is due to induced substance would be $\dpi{80} \frac{0.13n*0.73}{n} = 0.13*0.73 = 0.0949$ or 9.49%.

The event of failure due to induced substance would be the event of failure due to induced substance and outside factor. This can also be done as for we have $\dpi{80} P(OF) = 0.13$ (probability of outside factor) and $\dpi{80} P(IS|OF) = 0.73$ (probability of induced factor given failure is due to outside factor). Hence, as $\dpi{80} P(IS \cap OF) = P(IS | OF)*P(OF)$ , we have $\dpi{80} P(IS \cap OF) = 0.73*0.13 = 0.0949$ or 9.49%, which would be the probability of failure due to induced substances.

(b) Again, supposing that there are n occurrences of failure. Then, due to natural occurrences, there would be $\dpi{80} 0.87n$ events. Among these, there would be $\dpi{80} 0.87n*0.27$ cases due to disease and $\dpi{80} 0.87n*0.17$ due to infection. The probability that failure is due to disease would be $\dpi{80} \frac{0.87n*0.27}{n} = 0.2349$ , and probability that failure is due to infection would be $\dpi{80} \frac{0.87n*0.17}{n} = 0.1479$ . The probability that the failure is due to disease or infection would be $\dpi{80} 0.2349+0.1479=0.3828$ or 38.28%.

This can also be done as we have $\dpi{80} P(NO) = 0.87$ , $\dpi{80} P(D|NO) = 0.27$ and $\dpi{80} P(I|NO) = 0.17$ . Hence, we have $\dpi{80} P(D \cap NO) = 0.87 * 0.27$ and $\dpi{80} P(I \cap NO) = 0.87 * 0.17$ . The probability that failure is due to D or I would be $\dpi{80} P(D \cap NO) + P(I \cap NO) = 0.87 * 0.27 + 0.87 * 0.17 = 0.3828$ or 38.28%.