Question 1134020: a box of 80 candles consist of 30 defective and 50 non defective candles.If 10 of these candles are selected randomly what is the probability of all will be defective,6 will be non defective,all will be non defective?
Answer by Glaviolette(140)   (Show Source):
You can put this solution on YOUR website! 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
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