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Old 07-27-2007   #1
Justin T
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A light bulb manufacturer produces bulbs in batches of 50, and it is known from past experience that 5 of each batch will be defective If two bulbs are selected without replacement from a batch, what is the probability that both will be defective?(0.00816)Please show me how to do and the answer given is 0.00816.Thanks
 
Old 07-27-2007   #2
sniper4400
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it is a 4% chance that both will be defective
 
Old 07-27-2007   #3
odhumanities
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Starting "population" = 50 bulbsNumber of defective bulbs = 5 Probability of 1 bulb being defective = = Defective bulbs / Population=5/50 = .10 (10%)You selected one bulb that was defective. You didn't replace it in the population. So, repeating the process ...Your "new" population = 49Number of defective bulbs = 4REVISED Probability of 1 bulb being defective = = Defective bulbs / Population=4/49 = .08163 (8.16%)The probability of TWO bulbs being defective is the mathematical product of "P1" x "P2" (P = probability)So ...P1 x P2 == .10 x .08163= .008163--------------------------PS I actually like Steiner's approach (and Ksoileau's); Steiner's math made more intuitive sense to me and I fully understand the way the math replicates the reality. Though my math works (and is valid), it’s not immediately apparent (to me at least) why it works. I now understand the “x choose y” functionality (thank you Steiner and Ksoileau), with which I was not familiar, and thus had to work through it ... as follows ...Starting with the "50 Choose 2" “function” for all the unique pairs of two bulbs among the batch of 50:Start with Bulb1. That can be paired with Bulb2 through Bulb50. That's 49 unique pairs.Start with Bulb2. That can be paired with Bulb3 through Bulb50. It's already been paired with Bulb1 (immediately above), so we don't want to re-count that pair. So, that's 48 unique pairs.Start with Bulb3 ... 47 unique pairs.Start with Bulb4 ... 46 unique pairs.… etc., etc, ...Start with Bulb49 ... 1 unique pair.The total number of unique pairs is 1,225. 1,225 = (50x24) + 25[Rather than add 49+48+47+ … +3+2+1, I observe that I can: (i) start with the outermost elements in the series, i.e., 49+1, then (ii) add them and I get 50. Then 48+2=50, too. 47+3=50, etc., etc. That makes 24 paired integers, each of which totals 50 (24x50), plus the 25 that’s in the middle of the 49 elements in this additive series] … thus (50x24)+25=1,225].As for the five (5) DEFECTIVE bulbs, the "5 Choose 2""function" applies.Start with Defective1. That can be paired with Defective2 through Defective5. That’s 4 unique pairs.Start with Defective2. That can be paired with Defective3 through Defective5. (Note that it’s already been paired with Defective1, so you don’t want to recount that unique pair). That’s 3 unique pairs.Start with Defective3. That can be paired with Defective4 and Defective5. That’s 2 unique pairs.Finally, start with Defective4. That can only be paired with Defective5. That’s 1 unique pair.Total possible unique pairs of Defective Bulbs = 4+3+2+1 = 10.The probablity of selecting two defectives is [10 unique pairs of Defective Bulbs] DIVIDED by [1,225 unique pairs of All Bulbs] … 10/1,225 = 0.00816.
 
 

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