Question

A quality control technician at Cyberdyne Systems Corporation is checking each machine in
a row until he finds one that does not operate, but he will check no more than three machines,
even if they all operate. Assume each machine operates independently and has a probability
of operating of 0.9. What is the expected number of operating machines?

Answer 1

geometric distribution

Answer 2

i think, expected value = mean

Answer 3

No actually it's np

Answer 4

THis is confusing.

Answer 5

Sorry it's 1/p .

Answer 6

But if I do that I don't get the right answer.

Answer 7

P= P(3 work) + P(2 work, 1 fails) + P(1 works, 2 fails) + P(0 works, all 3 fail)
So P=0.82

Answer 8

1/p = 1.21 but the answer is 2.439

Answer 9

Interesting 2/0.82 = 2.439 but that makes no sense.

Answer 10

Are you saying it doesn't make sense because it's a decimal?

Answer 11

No no.

Answer 12

I'm saying it dsen't make sense because of that 2.

Answer 13

Besides for geometric distribution E(x) =1/p . That 2 dosent even come into play here.

Answer 14

And it should be geometric because it says until the first failure.

Answer 15

Have you tried using a calculator? A geometric distribution one, I mean.

Answer 16

We aren't allowed to use anything more sophisticated than a hand calculator.

Answer 17

Arragh. THis is frustrating me >.> .

Answer 18

A single mutiple choice should not be so hard >.< .

Answer 19

Well, the expected value would have to be between 0 and 1 at least.

Answer 20

No it has be be between 0 and 3.

Answer 21

SInce he checks 3 machines.

Answer 22

The probability that one fails is: \[
(1-0.9)+(0.9)^1(1-0.9) + (0.9)^2(1-0.9)
\]

Answer 23

\[
=0.271
\]

Answer 24

Oh wait, I see I misread it.

Answer 25

Regardless, we can still use \(1-0.271\) for the probability that none fail. This means \[
0\times (1-0.9)+1\times (0.9)^1(1-0.9) + 2\times (0.9)^2(1-0.9) + 3\times (1-0.271)
\]

Answer 26

I suggest using a table of values like
X |
-----------
f(x) |

Answer 27

\[
=2.439
\]

Answer 28

Dido, do you have an answer to check it?

Answer 29

Yeah that's right!

Answer 30

Well, it is hard to understand what I did?

Answer 31

A little.

Answer 32

The fallacy here is that the distribution at \(\Pr(X=3)\) is not geometric.

Answer 33

Why not? It says until the first one fails.

Answer 34

Doing so would be assuming that \(\Pr(X=4) \geq 0\). But how can that be possible if he would never check 4 of them?

Answer 35

Whoops, meant to pus \(\Pr(X=4)\gt 0\).

Answer 36

AHh right that makes a lot of sense yes.

Answer 37

Now explain what you did. It's P(3 fail)+P(2 fail)+P(1 fail)+P(0 fail) right?

Answer 38

I used geometric to calculate \(\Pr(X=k)\) when \(k=0,1,2\)

Answer 39

For \(\Pr(X=3)\), I used total probability: \[
1 = \Pr(X=0)+ \Pr(X=1)+ \Pr(X=2)+ \Pr(X=3)
\]

Answer 40

Once you calculate the probabilities, getting the mean shouldn't be an issue.

Answer 41

I know. It was just difficult to see this.

Answer 42

\[
\mu =\sum_{k=0}^{3}k\times \Pr(X=k)
\]

Answer 43

Thank you!

Answer 44

If you're still there @wio , why is the second term in your equation does not have a squared sign?

Answer 45

First term would have \((1-0.9)^0\), but I figured you wouldn't need it.

Answer 46

I mean would have \((0.9)^0\).

Answer 47

For 0 operating, 1 fail.
Second term is 1 operating 1 fail.

Answer 48

I had:

\[0(0.271)^3+1(0.90)(0.10)^2+2(0.90)^2(0.10)+3(0.90)^3\]

Answer 49

Should it not be 1 operation 2 fail?

Answer 50

operating*

Answer 51

Actually I guess it makes sense since he only needs 1 failure. SO we don't square it.

Answer 52

Right? @wio

Answer 53

Yeah.

Answer 54

Thanks :P .