Question

The probability is 1% that an electrical connector that is kept dry fails during the warranty period of a portable computer. If the connector is ever wet, the probability of a failure during the warranty period is 4%. If 90% of the connectors are kept dry and 10% are wet, what proportion of connectors fail during the warranty period? Round your answer to three decimal places (e.g. 98.765).

Answer 1

1.3%

Explanation:

Let A and B be the events

A = The connector fails during the warranty period given that it was kept dry

B = The connector fails during the warranty period given that it was ever wet

We are asked to find P(A or B) = P(A∪B)

But A and B are disjoints sets ( P(A∩B)=∅ ), so

P(A∪B) = P(A) + P(B).

On the other hand,

P(A) = 1% of 90% = 0.01*0.9 = 0.009

P(B) = 4% of 10% = 0.04*0.1 = 0.004

and

P(A) + P(B) = 0.009 + 0.004 = 0.013 = 1.3%

So, the proportion of connectors that might fail during the warranty period is 1.3%