Question

A construction company employs two sales engineers.
Engineer 1 does the work of estimating cost for 30% of jobs bid by the company.
Engineer 2 does the work for 70% of jobs bid by the company. It is known that
the error rate for engineer 1 is such that 0.04 is the probability of an error when
he does the work, whereas the probability of an error in the work of engineer 2 is
0.02. Suppose a bid arrives and a serious error occurs in estimating cost. Which
engineer would you guess did the work? Explain and show all your calculation.

Answer 1

Answer:

your question is very good

Answer 2

Explanation:

Answer:

Engineer 1

Step-by-step explanation:

Based on the information given:

P( 1 ) = 0.7

P( 2 ) = 0.3

P(A | B) is a conditional probability: the likelihood of event A occurring given that B is true.

P( E | 1 ) = 0.02

P( E | 2 ) = 0.04

Then P( E | 1 ) is the probability that an error occurs when engineer 1 does the work and P( E | 2 ) is the probability that an error occurs when engineer 2 does the work.

To have an idea of which engineer is more likely to do the work when an error occurs you need to calculate P( 1 | E ) and P( 2 | E ), The probability that engineer 1 does the work when an error occurs and the probability that engineer 2 does the work when an error occurs.

The Bayes's theorem states:

P( B | A ) = \frac{P( A | B )*P( A )}{P( B )}P(B∣A)=

P(B)

P(A∣B)∗P(A)

Using the notation above:

P( 1 | E ) = \frac{P( E | 1 )*P( 1 )}{P( E )}P(1∣E)=

P(E)

P(E∣1)∗P(1)

P( E ) = 0.7*0.02 + 0.3*0.04 = 0.026 /// the probability that engineer 1 does the work and an error occurs or the probability that engineer 2 does the work and an error ocurrs.

P( 1 | E ) = \frac{0.04*0.3}{0.026}=0.538P(1∣E)=

0.026

0.04∗0.3

=0.538

Doing the same for engineer 2:

P( 2 | E ) = \frac{0.04*0.3}{0.026}=0.462P(2∣E)=

0.026

0.04∗0.3

=0.462