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.

Note that the broken chair can be chosen either:

*

*first with probability: $$P_1 = \frac {2}{50}\times \frac {48}{49}\times \frac {47}{48}\times \frac {46}{47}\times \frac {45}{46} $$


*second with probability: $$P_2 = \frac {48}{50}\times \frac {2}{49}\times \frac {47}{48}\times \frac {46}{47}\times \frac {45}{46}$$
Surely you can complete the other cases.
