I cannot get to the solution 1.02 and I am way off for the following question. A

Question

I cannot get to the solution 1.02 and I am way off for the following question.

A light bulb manufacturer introduces a new product. The firm believes that there is a(n) 88 percent of the bulbs functioning as advertised. The firm's engineering department tests a random sample of bulbs. In the past, testing returned a positive result 94 percent of the time when bulbs actually functioned properly (as advertised). Further, if the bulbs failed to work as advertised, testing returned a negative result 97 percent of the time. If the bulb works as advertised, the firm will earn $4.8 in profit. If the fail to work as advertised, the firm will lose $0.7. Calculate the expected value of the bulb given a negative test result. Round your answer to two decimal places.

Explanation / Answer

Here, we are given that:

P( function ) = 0.88, therefore P( not function ) = 0.12

P( positive | function ) = 0.94 and P( negative | not function ) = 0.97

Therefore, P( negative | function ) = 1 - 0.94 = 0.06

Using law of total probability, we get:

P( negative ) = P( negative | function )P( function ) + P( negative | not function )P( not function )

P( negative ) = 0.06*0.88 + 0.97*0.12 = 0.1692

Now using bayes theorem, we get:

P( function | negative ) = P( negative | function )P( function ) / P( negative ) = 0.06*0.88 / 0.1692 = 0.3121
P( not function | negative ) = P( negative | not function )P( not function ) / P( negative ) = 0.97*0.12 /0.1692 = 0.6879

Therefore expected value of the bulb here is computed as:

= 4.8*P( function | negative ) - 0.7*P( not function | negative )

= 4.8*0.3121 - 0.7*0.6879

= 1.02

Therefore 1.02 is the expected value here.

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