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The probability that coal fired electric generator 1 is working is .10. The probability that coal fired electric generator 2 is working is .20. What are the probabilities that: Both work? Neither work? Only one works?

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impossible to answer since nothing is noted about the dependence or independence between the two generators. –  Ittay Weiss Feb 9 '13 at 22:11
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... and in practice the two probabilities are certainly not independent, since there are underlying causes that would tend to apply to both generators that the same time (coal shortages, strikes, earthquakes, etc) and it would be a fantastic coincidence if these were canceled out exactly by causes where there is a dependence that biases against simultaneous failures. –  Henning Makholm Feb 9 '13 at 22:24
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HINTS: Assuming that the generators are completely independent, the probability that both work is the product of the individual probabilities of their working, or $0.10\cdot0.20=0.02$. (Without that some assumption about dependence or independence of the generators the questions cannot be answered; unless the point of the question is to see whether you understand that this information is necessary, you’re probably expected to assume independence of the generators, and I will do so.)

The probability that the first generator does not work is $1.00-0.10=0.90$. What is the probability that the second generator does not work? What is the probability that both fail to work? This problem is now just like the first one.

If you’ve got this far, you know the probability that both work and the probability that both fail to work. The event that exactly one of them works covers all other outcomes, and the probabilities of the possible outcomes must add up to $1$, so the probability that exactly one works is ... ?

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so would the answer to, if neither works?, be 0.72 –  Natasha Feb 9 '13 at 22:39
    
@Natasha: Yes, it would. –  Brian M. Scott Feb 9 '13 at 22:40
    
and only one works?, 0.26 –  Natasha Feb 9 '13 at 22:40
    
@Natasha: That’s right. –  Brian M. Scott Feb 9 '13 at 22:40
    
thank you @Brian M. Scott you are a life saver –  Natasha Feb 9 '13 at 22:42
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My 8th grader writes thusly:

"Even though there is not enough information provided, Assuming that there are no dependencies: $(1/10)*(2/10) = (2/100)$ so the probability of them both working is 2 out of 100. $(9/10)*(8/10) = (72/100)$ so the probability of neither working is 72 out of 100, or 36 out of 50, or 18 out of 25. $(1/10)*(8/10) = (8/100)$ and $(9/10)*(2/10) = 18/100$ so the final product should be $(8/100)+(18/100)=(26/100)$"

(Please do not upvote or downvote; it was his work. If inappropriate, I will delete.)

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Tell him that he nailed it. –  Ross Millikan Feb 9 '13 at 22:52
    
I just told him that a guy with a huge reputation score said he nailed it, he beamed. –  Ron Gordon Feb 9 '13 at 22:54
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Hint: The chance that a given one is not working is 1 minus the chance that it is working. If you assume that the events of each working are independent, the probability they both work is the product of the probability of each.

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