probability toys are defective.

[galactus]

Here's a probability I am not sure how to tackle. I seen this in an old book.

Suppose a toy manufacturer has found from past checks that the probability of defects in a sample of 4 different toys are
0 are defects is .60
1 is a defect is .10
2 are defects is .04
3 are defects is .03
4 are defects are .23

Suppose one is chosen at random and it's a defect. What's the probability
that the all are defects?.

I know this is probably something simple, but I have a mental block.

 

[pka]

Suppose a toy manufacturer has found from past checks that the probability of defects in a sample of 4 different toys are
0 are defects is .60
1 is a defect is .10
2 are defects is .04
3 are defects is .03
4 are defects are .10
.......................0.87

What are to make of the missing probability?

 

[galactus]

I'm sorry, pka. That was a stupid typo.

 

[pka]

Let’s say that R stands for the event that the randomly chosen toy, from the four, is defective and X is the number of defective toys in the sample.
The question being asked is \(\displaystyle P(X=4|R)\).
\(\displaystyle \begin{array}{rcl}
P(R) & = & \sum\limits_{k = 0}^4 {P\left( {R \cap \left[ {X = k} \right]} \right)} \\
& = & \sum\limits_{k = 0}^4 {P\left( {R|\left[ {X = k} \right]} \right)P\left( {\left[ {X = k} \right]} \right)} \\
\end{array}\)
What is next?

 

[galactus]

Well, let's see, pka. I hope I understand you correctly.

The prob. the randomy chosen is a defect given 0 are defects would be

(0)(.6)

The prob. the randomy chosen is a defect given 1 is a defect would be:

(1/4)(0.1)

The prob, the randomy chosen is a defect given 2 are defects would be:

(2/4)(.04)

The prob. the randomy chosen is a defect given 3 are defects would be:

(3/4)(.03)

And, the prob. the randomy chosen is a defect given 4 are defects would be (1)(.23)

Adding them up I get 29.75% the prob they're all defects given the randomy chosen was a defect.

I have a feeling I did something wrong. I have not encountered a problem like this before. I thought it was interesting.

 

[pka]

Well you did find P(R).
That is the probability that choosing one from the sample of four and finding it defecttive.

But the question is to find P(X=4|R).

 

[galactus]

I am sorry to say, pka, I have no idea then. The probability all 4 are defects given a chosen one is defective?. Hmmmmm.
I tried all sorts of scenarios, but I can't see how to set it up. It's probably obvious, but I don't see it. The way I had it seemed logical, but apparently not. Oh well, thanks for your help.

 

[pka]

Look up Bayes’ Theorem in a probability text.
This is a typical such problem concerning partitions of the space.
In this case there are five cells.
\(\displaystyle \begin{array}{rcl}
P\left( {X = 4|R} \right) & = & \frac{{P\left( {X = 4 \cap R} \right)}}{{P(R)}} \\
& = & \frac{{P\left( {R|X = 4} \right)P(X = 4)}}{{P(R)}} \\
\end{array}\)

 

[galactus]

I actually tried Bayes' theorem. That's what came to mind first. It turned cumbersome, so I quit. I reckon this shows one shouldn't be a quitter.
Thank you.

 

[pka]

I actually tried Bayes' theorem. It turned cumbersome, so I quit.

I tell students that as a rule of thumb start with the denominator.
In this case P(R).
The numerator will be part of that calculation.

 

[galactus]

Okey-doke. Believe it or not, I teach Elements of Statistics at a local community college, including Bayes' theorem. But the probability in that course is just beginning stuff. Probability and counting is one of my favorite areas of mathematics. But, I must admit, it sure burns my synapses sometimes.

I got it(I think) . Thanks.

\(\displaystyle \frac{(1)(\frac{23}{100})}{\frac{1}{4}(\frac{1}{10})+\frac{2}{4}(\frac{1}{25})+\frac{3}{4}(\frac{3}{100})+\frac{4}{4}(\frac{23}{100})}\)

\(\displaystyle =\frac{92}{119}\approx{.773}\)



You learned me something pka. I had never used Bayes' theorem in that light before. That is, summing up the different cases because we had 5 different probs. It was apparent once you pointed me in the right direction with Bayes' theorem.