1. ## simple i think

We have 60 TV's on the shop , 5 of them have defects.
if we buy 2 TV 's how is the probability that both have defects ?

2. Yes, simply, prob to get 1 defective tv =5/60

Ermm my mistake, binomial distribution applies here....

http://personal.maths.surrey.ac.uk/s...poiss_bin.html

Some hints there

3. Originally Posted by beq!x
We have 60 TV's on the shop , 5 of them have defects.
if we buy 2 TV 's how is the probability that both have defects ?
You assume that the 60 TV's in the shop are typical so that the proportion of defectives is 1/12, and the two TV's bought are unrelated in the defectivness department.

Then the probability that the two TV are both defective is then (1/12)(1/12).

CB

4. Yes, gave the same answer as CB, then totally gone mad and deleted it...

5. what is CB ?

6. Originally Posted by beq!x
what is CB ?
It is CaptainBlack's initials.

CB

7. Originally Posted by CaptainBlack
You assume that the 60 TV's in the shop are typical so that the proportion of defectives is 1/12, and the two TV's bought are unrelated in the defectivness department.

Then the probability that the two TV are both defective is then (1/12)(1/12).

CB
But the chance of you buying one defective TV is 5/60.
Then if you buy another tv, the chances of it being defective are 4/59.
5/60*4/59= 20/3540 = 1/177...

You're assuming independence, are you not?
Which they're not...

8. but what if i buy 2 TV's in the same time ?

9. Originally Posted by Unenlightened
But the chance of you buying one defective TV is 5/60.
Then if you buy another tv, the chances of it being defective are 4/59.
5/60*4/59= 20/3540 = 1/177...

You're assuming independence, are you not?
Which they're not...
That's not the way I read the question. The way I read it is we have 60 TV's in our shop 5 defective, we buy two more, what is the probability that these two new stock TV's are defective.

Now if the two "we" 's in the question refer to different actors then yes the answer is different.

CB

10. Originally Posted by Unenlightened
But the chance of you buying one defective TV is 5/60.
Then if you buy another tv, the chances of it being defective are 4/59.
5/60*4/59= 20/3540 = 1/177...

You're assuming independence, are you not?
Which they're not...
so this is the right answer ?

11. Originally Posted by CaptainBlack
That's not the way I read the question. The way I read it is we have 60 TV's in our shop 5 defective, we buy two more, what is the probability that these two new stock TV's are defective.

Now if the two "we" 's in the question refer to different actors then yes the answer is different.

CB
Ah..... that's not how I read it...
I see..

Originally Posted by beq!x
so this is the right answer ?
That would depend, on what the question is...

12. Originally Posted by beq!x
We have 60 TV's on the shop , 5 of them have defects.
if we buy 2 TV 's how is the probability that both have defects ?
My interpretation is that you're buying two TV's from the pool of 60 TV's.

So the probability of both being defective will be $\frac{^5C_2 \cdot ^{55}C_0}{^{60}C_2}$. You can think about where the different bits of this answer come from, and I'll let you simplify it.

Alternatively, you can use the hypergeometric distribution if you know it.

Originally Posted by beq!x
what is CB ?
Originally Posted by CaptainBlack
It is CaptainBlack's initials.

CB
And I thought it stood for Citizen Band. Or was that Citizen Banned .... ?