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horrorfan [7]
3 years ago
11

9) A construction company employs 2 sales engineers. Engineer 1 does the work in estimating cost for 70% of jobs bid by the comp

any. Engineer 2 does the work for 30% of jobs bid by the company. It is known that the error rate for Engineer 1 is such that 0.02 is the probability of an error when he does the work, whereas the probability of an error in the work of Engineer 2 is 0.04. What is the probability that a bid provided by the company will contain an error in the cost estimates?
If a serious error is known occurred, what is the probability that it was made by Engineer 2?
Engineering
1 answer:
Irina18 [472]3 years ago
3 0

The probability that the error occurred when Engineer 2 made the mistake is 0.462 on the other hand the probability that the error occurred when the engineer 1 made the mistake is 0.538

Explanation:

Let E_{1} denote the event that the 1st  engineer  does the work.so we write

P(E)_{1}=0.7

Let E_{2} denote the event that the 2nd engineer  does the work .So we write

P(E)_{2}=0.3

Let O denote the event during which the error occurred .so we write

P(O/E_{1} )=0.02(GIVEN)

P(O/E_{2} )=0.04(GIVEN)

  • The probability that the error occurred when the first engineer performed the work is P(E_{1} /O)
  • The probability that the error occurred when the first engineer performed the work is P(E_{2} /O)

Now we need to find when did the error in the work occur so we will compare the probability of the work done by <u>engineer 1 </u>and <u>engineer 2 </u>

<u></u>

<u>lets find the Probability of the Engineer 1</u>

<u>Using Bayes theorem,we get</u>

<u></u>

<u></u>P(E_{1} /O) =0.02*0.7/0.02*0.7+0.04*0.3 = 0.014/0.026=0.538

<u>lets find the Probability of the Engineer 2</u>

<u></u>P(E_{2} /O) =0.04*0.3/0.02*0.7+0.04*0.3=0.012/0.026=0.462

Since ,0.462<0.538 so it is more prominent that the Engineer 1 did the work when the error occurred

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