probability toys are defective.

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.

galactus said:

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

Click to expand...

What are to make of the missing probability?

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

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?

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.

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).

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.

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.
\(\displaystyle \begin{array}{rcl}
P\left( {X = 4|R} \right) & = & \frac{{P\left( {X = 4 \cap R} \right)}}{{P(R)}} \\
& = & \frac{{P\left( {R|X = 4} \right)P(X = 4)}}{{P(R)}} \\
\end{array}\)

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.

galactus said:

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

Click to expand...

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.

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.