Question 934837: In a shipment of 2,500 video games, 97 are defective. If a person buys two games, what is the probability that only one is defective?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 2500 video games.
97 are defective
2403 are good.
probability of exactly one defective and one good would be equal to c(97,1) * c(2403,1) / (c(2500,2) which is equal to 233091 / 3123750 which is equal to .0746189676.
this is equal to the number of ways you can get 1 defective out of 97 defectives times the number of ways you can get 1 good out of 2403 good, all divided by the number of ways you can get 2 out of 2500 regardless of whether they are defective or not.
c(97,1) is the combination formula of 97! / (1! * 96!) which is equal to (97 * 96!) / (1! * 96!) which is equal to 97 / 1 which is equal to 97.
c(2403,1) is the combination formula of 2403! / (1! * 2402!) which is equal to (2403 * 2402!) / (1! * 2402!) which is equal to 2403 / 1 which is equal to 2403.
c(2500,2) is the combination formula of 2500! / (2! * 2488!) which is equal to (2500 * 2499 * 2498!) / (2! * 2498!) which is equal to (2500 * 2499) / 2 which is equal to 3123750.
you wind up with (97 * 2403) / 3123750) which is equal to .0746189676
you can also look at it as follows:
probability of one defective is 97/2500
probability of one good is 2403/2500
the probability of 1 defective then one good without replacement is 97/2500 * 2403/2499.
the probability of 1 good then one defective without replacement is 2403/2500 * 97/2499.
the probability of one defective and one good is the probability of one defective then one good plus the probability of one good then one defective which is equal to 97/2500 * 2403/2499 + 2403/2500 + 97/2499.
97/2500 * 2403/2499 = .0373094838
2403/2500 * 97/2499 = .0373094838
add them up and you get .0746189676
the general form of the combination formula is:
c(n,x) = n! / (x! * (n-x)!)
c(n,x) can also be shown as nCx.
the formula is the same.
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