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horrorfan [7]
4 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]4 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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Answer:

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Explanation:

given data

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average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

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put here value

Δp = \frac{2*2.69*10^{-2}}{100*10^{-6}}  

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

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