Question 1134020
Since there are 30 defective out of 80, the probability of the first one being defective is 30/80.  Then there is one less defective one and one less candle, so the next probability is 29/79, and so on.
P(all 10 defective) = 30/80 * 29/79 * 28/78 * 27/77 * 26/76 * 25/75 * 24/74 * 23/73 * 22/72 * 21/71 = 1.02 x 10^-5

P(6 will be non-defective) is a little trickier to do because of the number of different ways that 6 out of 10 can happen.  So, I'll start by treating it as the first 6 being defective and the last 4 are not.
= 30/80 * 29/79 * 28/78 * 27/77 * 26/76 * 25/75 * 50/74 * 49/73 * 48/72 * 47/71 = 3.31 * 10 ^-4

Using combinations, there are 210 different ways that 6 out of 10 candles can be the defective ones.  Therefore, I would multiply the above probability by 210 to get .0695

Lastly, non defective would be much like the process for all defective but using 50/80 then 49/79, and so on.
P(non defective) = 50/80 * 49/79 * 48/78 * 47/77 * 46/76 * 45/75 * 44/74 * 43/73 * 42/72 * 41/71 = .0062