A factory producing electric motors has 0.2% defective rate.

A factory producing electric motors has 0.2% defective rate. A quality control inspector tests randomly selected motors from this production line.
a. what is the probability that the first defective motor will the the sixth one?
b. what is the probability that the first defective motor will be among the sixth tested?
c. What is the expected waiting time before the first defective motor is tested?

You should recognize this for the specific distribution that can be used to model it.

Generally, just add it up...

a) Pr(good)*Pr(good)*Pr(good)*Pr(good)*Pr(good)*Pr(bad) = 0.998^5*0.002

b) First 0.002
Second 0.998*0.002
Third 0.998^2*0.002

or Notice the Binomial version Pr(at least one defective in six tries) = 1 - Pr(no defectives in six tries) = 1 - 0.998^6

This one may be a bit confusing if you don't recognize that there is no requirement that only one defective occurs. You could get six straight defectives.

c) Waiting time is a little tricky. If the first one is defective, the waiting time may be zero. It depends on how you are thinking about the index.

0*0.002 + 1*0.998*0.002 + 2*0.998^2*0.002 + 3*0.998^3*0.002 + 4*0.998^4*0.002 + 5*0.998^5*0.002 + ...

Can you add that?

Like I said in the beginning, if you recognize the convenient standard distribution, it's all much simpler.

Re: A factory producing electric motors has 0.2% defective r

Hello, kimmy_koo51!

A factory producing electric motors has 0.2% defective rate.
A quality control inspector tests randomly selected motors from this production line.

Click to expand...

Let: \(\displaystyle \,d\:=\(\text{d{e}f}) \:=\:0.002\)
Let: \(\displaystyle \,g \:=\(\text{good}) \:=\:0.998\)

a. What is the probability that the first defective motor will the the sixth one?

Click to expand...

The first five must be good and the sixth must be defective: \(\displaystyle \:g\,g\,g\,g\,g\,d\)

Answer: \(\displaystyle \(\text{6th def})\:=\:g^5d \:=\0.998)^5(0.002) \:=\:0.00198008\)

b. What is the probability that the first defective motor will be among the sixth tested?

Click to expand...

\(\displaystyle \begin{array}{cccc}1^{st}\text{ def:}\: & d & \,=\, & 0.002 \\
2^{nd}\text{ def:} & g\,d & = & (0.998)(0.002) \\
3^{rd}\text{ def:} & g\,g\,d & = & (0.998)^2(0.002) \\
4^{th}\text{ def:} & g\,g\,g\,d & = & (0.998)^3(0.002) \\
5^{th}\text{ def:} & g\,g\,g\,g\,d & = & (0.998)^4(0.002) \\
6^{th}\text{ def:} & g\,g\,g\,g\,g\,d & = & (0.998)^5(0.002)
\end{array}\)

. . . and add them up.

c. What is the expected waiting time before the first defective motor is tested?

Click to expand...

There is a formula for this question, but I'll do it out the long way . . .


The expected time is: \(\displaystyle \:E \:=\:1\cdot d\,+\,2\cdot gd \,+\,3\cdot g^2d \,+\,4\cdot g^3d \,+\,5\cdot g^4d\,+\,\cdots\)

Multiply by \(\displaystyle g:\) . . . . . .\(\displaystyle gE \:=\) . . . . . .\(\displaystyle gd\,+\,2\cdot g^2d\,+\,3\cdot g^3d\,+\,4\cdot g^4d\,+\,\cdots\)

Subtract: . . . . \(\displaystyle (1\,-\,g)E\:=\:\;d\;+\;\;gd\;+\;g^2d\;+\;g^3d\;+\;g^4d\,+\,\cdots\)

. . The right side is a geometric series with first term \(\displaystyle d\) and common ratio \(\displaystyle g\).
. . . . Its sum is: \(\displaystyle \:\frac{d}{1\,-\,g}\)

Hence, we have: \(\displaystyle \1\,-\,g)E\:=\:\frac{d}{1\,-\,g}\;\;\Rightarrow\;\;E \:=\:\frac{d}{(1\,-\,g)^2}\)

Therefore: \(\displaystyle \:E \;=\;\frac{0.002}{(1\,-\,0.998)^2}\;=\;\frac{0.002}{0.002^2}\;=\;\frac{1}{0.002} \;=\;500\)