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 ?
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
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
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 $\displaystyle \frac{^5C_2 \cdot ^{55}C_0}{^{60}C_2}$. 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.
And I thought it stood for Citizen Band. Or was that Citizen Banned .... ?