Question

Errors in estimating job costs. A construction companyemploys three sales engineers. Engineers 1, 2, and 3 estimate the costs of$\mathbf{30}\mathbf{\%}$,$\mathbf{70}\mathbf{\%}$, and$\mathbf{50}\mathbf{\%}$, respectively, of all jobs bid by the company. For$\mathbf{i}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}$, define${\mathbf{E}}_{\mathbf{i}}$to be the event that a job is estimated by engineer i.The following probabilities describe the rates at which the engineers make serious errors in estimating costs:

$\mathbf{P}\mathbf{\left(}\mathbf{error}\mathbf{\right|}{\mathbf{E}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{error}\mathbf{\right|}{\mathbf{E}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{03}\mathbf{,}\mathbf{andP}\mathbf{\left(}\mathbf{error}\mathbf{\right|}{\mathbf{E}}_{\mathbf{3}}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{.}$

1. If a particular bid results in a serious error in estimating job cost, what is the probability that the error was made by engineer 1?
2. If a particular bid results in a serious error in estimating job cost, what is the probability that the error was made by engineer 2?
3. If a particular bid results in a serious error in estimating job cost, what is the probability that the error was made by engineer 3?
4. Based on the probabilities, parts a–c, which engineer is most likely responsible for making the serious error?

Short answer

The probabilities of all parts are

1. By engineer 1 is 0.1579.
2. By engineer 2 is 0.6.
3. By engineer 3 is 0.1
4. The engineer 2 is most likely responsible for making the serious error.

Step-by-step solution

Important formula

The Baye’s formula is

$\begin{array}{rcl}\mathbf{P}\left(\frac{{\mathbf{B}}_{\mathbf{i}}}{\mathbf{A}}\right)& \mathbf{=}& \frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}\cap \mathbf{A}\mathbf{)}}{\mathbf{P}\mathbf{(}\mathbf{A})}\\ & \mathbf{=}& \frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{i}}}\right)}{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{1}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{1}}}\right)\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{2}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\right)\mathbf{+}...\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{k}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{k}}}\right)}\\ & & \end{array}$

(a) Step 2: The probability that the error was made by engineer 1.

Given

$P\left(error|E_1\right)=0.01,P\left(error|E_2\right)=0.03,\text{and} P\left(error|E_3\right)=0.02$

$P\left(E_1\right)=0.30,P\left(E_2\right)=0.20,P\left(E_3\right)=0.50$

$\begin{array}{rcl}P\left(error|E_1\right)& =& \frac{P\left(E_3\right)P\left(\frac{errror}{E_1}\right)}{P\left(E_1\right)P\left(\frac{error}{E_1}\right)+P\left(E_2\right)P\left(\frac{error}{E_2}\right)+P\left(E_3\right)P\left(\frac{error}{E_3}\right)}\\ & =& \frac{\left(0.30\right)\left(0.01\right)}{\left(0.30\right)\left(0.01\right)+\left(0.20\right)\left(0.03\right)+\left(0.50\right)\left(0.02\right)}\\ & =& 0.1579\\ & & \end{array}$

So, by engineer 1 is 0.1579.

(b) Step 3: Evaluate the probability that the error was made by engineer 2.

$\begin{array}{rcl}P\left(error|E_2\right)& =& \frac{P\left(E_3\right)P\left(\frac{errror}{E_1}\right)}{P\left(E_1\right)P\left(\frac{error}{E_1}\right)+P\left(E_2\right)P\left(\frac{error}{E_2}\right)+P\left(E_3\right)P\left(\frac{error}{E_3}\right)}\\ & =& \frac{\left(0.20\right)\left(0.03\right)}{\left(0.30\right)\left(0.01\right)+\left(0.20\right)\left(0.03\right)+\left(0.50\right)\left(0.02\right)}\\ & =& 0.6\\ & & \end{array}$

Hence, by engineer 2 is 0.6.

(c) Step 4: Determine the probability that the error was made by engineer 3

$\begin{array}{rcl}P\left(error|E_2\right)& =& \frac{P\left(E_3\right)P\left(\frac{errror}{E_1}\right)}{P\left(E_1\right)P\left(\frac{error}{E_1}\right)+P\left(E_2\right)P\left(\frac{error}{E_2}\right)+P\left(E_3\right)P\left(\frac{error}{E_3}\right)}\\ & =& \frac{\left(0.50\right)\left(0.02\right)}{\left(0.30\right)\left(0.01\right)+\left(0.20\right)\left(0.03\right)+\left(0.50\right)\left(0.02\right)}\\ & =& 0.1\\ & & \end{array}$

So, by engineer 3 is 0.1

(d) Step 5: which engineer is most likely responsible for making the serious error

According to the results of the part a to c, it is clear that the engineer 2 is most likely responsible for making the serious error.

Therefore, the significant error was probably made by engineer 2 (most likely).