Calculating probability using combinations We have 50 chairs. 2 out of them are broken. What is probability that out of 5 taken chairs 4 or 5 are not broken?
We can calculate it by using $P(4 not broken) + P( 5 not broken)$
$P( 5 not broken ) = \frac{48}{50} * \frac{47}{49} * \frac{46}{48} * \frac{45}{47}* \frac{44}{46}$
But what about $P(4 not broken)$? If i take one chair probability that another chair will be broken/ not be broken is reduced. In this case can i use
$P(4 not broken) = C(5,4)* (\frac{48}{50} * \frac{47}{49} * \frac{46}{48} * \frac{45}{47})^{4} * \frac{2}{46}$ ?
Do i need to calcualte it using combinations?
Thanks for answers.

I would calculate it this way:
Since it's without replacement and without order:
$P(A)=\frac{\binom {48} {k}\binom {2} {k}}{\binom {50} {5}}$
Insert K value 4 and 5 and your solutions are:
$P(4 not brokenchairs)=\frac{\binom {48} {4}\binom {2} {1}}{\binom {50} {5}}=\frac{389160}{2118760}=\frac{9}{49}=.18367$
$P(5 not brokenchairs)=\frac{\binom {48} {5}\binom {2} {0}}{\binom {50} {5}}=\frac{1712304}{2118760}=\frac{198}{245}=.80816$
