Probability Question

If you picked a sample of 1 with a probability of defects being p (any p whatsoever), what would the probability of picking a defect be?

If I picked a sample of 1 with the probablity of picking a defect being p=4, it would be 4/11. Is this right?

Okay...thinking of only the second half of the priginal problem. There would be a probability of 1/3 chance in picking a defect. Right?

Gr8fu13 said:

In a shipment of 11 CD players, two are defective. Find the probability that a sample of 3 CD players will contain no more than one defective one.

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This is basically a counting question.
How many ways are there to select three of those eleven players?
How many ways are there to select three of those eleven players which are not defected ?
How many ways are there to select three of those eleven players where exactly one is defective?

To find the probability that a sample of 3 CD players will contain no more than one defective one, add the last two and divide by the first.

pka said:

This is basically a counting question.
How many ways are there to select three of those eleven players?
Would this just be 3x11= 33 different ways?

How many ways are there to select three of those eleven players which are not defected ?
Would this just be the amount of non defective players times 3? 7x3=24 different ways?

How many ways are there to select three of those eleven players where exactly one is defective?
I am not sure about this one.

To find the probability that a sample of 3 CD players will contain no more than one defective one, add the last two and divide by the first.

Click to expand...

Am I even close?

Gr8fu13 said:

Am I even close?

Click to expand...

Have you studied any thing about combinations?
If we have \(\displaystyle N\) different items and \(\displaystyle 1\le K \le N\) the number of ways to select \(\displaystyle K\) items from the \(\displaystyle N\) is a combination.

There are many notations for that.
Here are three: \(\displaystyle ^N\mathcal{C}_K,~\mathcal{C}(N,K)~,~\binom{N}{K}\).

Most mathematicians prefer the last one.
So \(\displaystyle \binom{N}{K}=\frac{N!}{(N-K)!(K!)}\)

So to choose three players from eleven, there are
\(\displaystyle \binom{11}{3}=\frac{11!}{(8!)!(3!)}=165\) ways to do it.

There are \(\displaystyle \binom{8}{3}\) ways to choose three that are not defective.

Can you finish?

How about this:

(11/8) = 11!/(8!)!(3!) = 165 ways to choose 3 players
(8/3) ways to choose three not defective
(9/165) ways to choose where exactly one is defective
1 - 9/165 = 156/165 or 52/55 simplified probability of no more than one defective unit in a sample of 3.

Am I atleast getting closer?
This is so frustrating

Sorry I thought there were three defected units not just two.
So the correct answer is:
\(\displaystyle \dfrac{\binom{9}{3}+2\binom{9}{2}}{\binom{11}{3}}\)