SOLUTION: A bag contains 20 batteries, of which six are defective. Selecting two at random, without replacement, determine the probability that none of the batteries you select are good. >3

Algebra ->  Probability-and-statistics -> SOLUTION: A bag contains 20 batteries, of which six are defective. Selecting two at random, without replacement, determine the probability that none of the batteries you select are good. >3      Log On


   

Question 460728: A bag contains 20 batteries, of which six are defective. Selecting two at random, without replacement, determine the probability that none of the batteries you select are good.
>3/5
>3/10
>3/38
>9/100
please show me how you get the answer?
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
P(2 Defective) = P(Defective AND defective)


P(2 Defective) = P(Defective) * P(Defective)


P(2 Defective) = (6/20)*(5/19)


P(2 Defective) = (6*5)/(20*19)


P(2 Defective) = 30/380


P(2 Defective) = 3/38


So the probability that both are defective is 3/38, which means that the answer is choice C)

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A bag contains 20 batteries, of which six are defective. Selecting two at random, without replacement, determine the probability that none of the batteries you select are good.
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# of ways to select 2 defective batteries: 6C2 = 15
# of ways to select 2 batteries from 20: 20C2 = 190
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P(select 2 defective) = 15/190 = 3/38
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Cheers,
Stan H.
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>3/5
>3/10
>3/38
>9/100