Question

The probability s 1% that an electrical connector that is kept d y fails dur ng the warranty period of ortable compute lf the connector ıs eve wet kept dry and 10% are wet, what proportion of connectors fall during the wamanty period? Round your answer to three decimal places (e.g. 98.765). e proba bility of a failure during the warranty po od is 90% of the connectors are exact number, no tolerance

Answer 1

Let D be the event that the connector is dry and W be the event that the connector is wet.

F be the event that the portable computer fails during the warranty period.

Given P(F/D) = 1/100 = 0.01, P(F/W) = 9/100 = 0.09, P(D) = 90/100 = 0.9, P(W) = 10/100 = 0.1

Total probaility theorem - Let $\small A_{i}$ form the sample space S then for an event B, $\small P(B) = \sum P(A_{i})*P(B/A_{i})$ .

$\small \Rightarrow$ P(F) = P(D)*P(F/D) + P(W)*P(F/W) = 0.9*0.01 + 0.1*0.09 = 0.018

Therefore the probability that the portable computer fails during the warranty period is 0.018