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Probability question

by MTariq » Tue Mar 16, 2010 9:24 am
Can someone point out the mistake. what am I doing wronge? Answer should be the same using both approachesA machine is made up of two components, A and B. Each component either works or fails. The failure or non-failure of one component is independent of the failure or non-failure of the other component. The machine works if at least one of the components works. If the probability that each component works is 2/3, what is the probability that the machine works?

Prob that neither component works = 1/3*1/3
Prob that machine works = 1- 1/9 = 8/9

Prob Machine works = Prob (A works) * Prob (B does not work) + Prob (B works) * Prob (A does not work)
= 2/3*1/3 + 2/3*1/3
= 2/9 + 2/9 = 4/9
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by kstv » Tue Mar 16, 2010 9:48 am
So prob of the machine working is when at least one of the comp. works. But what about the probability that both comp. might work. U are not accounting for it in the second method.

while it is being assumed in your calculation in the first method. Probability of both working is 2/3*2/3 = 4/9 the diff in the two methods.


think of components as switches it is either on or off.
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by MTariq » Tue Mar 16, 2010 10:11 am
Ohhkay, I was missing out the third scenerio.
Got it, Thanks!
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by hooliganpete » Fri Mar 19, 2010 11:14 am
The reason above seems incorrect to me but maybe someone can elaborate. If the machine can function with only one component functioning, then the only condition that would render the machine inoperable would be when neither components A nor B work.

The probability of either not working is 1/3. Thus, the probability that neither works is (1/3)(1/3)=1/9. Probability of the machine functioning minus the only condition which makes it not function is 1-(1/9)=8/9.

Can someone explain why, if at all, the above reasoning is incorrect?
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by harshavardhanc » Fri Mar 19, 2010 12:11 pm
hooliganpete wrote:The reason above seems incorrect to me but maybe someone can elaborate. If the machine can function with only one component functioning, then the only condition that would render the machine inoperable would be when neither components A nor B work.

The probability of either not working is 1/3. Thus, the probability that neither works is (1/3)(1/3)=1/9. Probability of the machine functioning minus the only condition which makes it not function is 1-(1/9)=8/9.

Can someone explain why, if at all, the above reasoning is incorrect?

Pete,

Your reasoning is the first method employed by MTariq, which is one of the correct ways to reach the answer.

But, he was not getting the same answer in the second method, as he didn't consider the case where both the machines work. That's what kstv explained in his post. That probability will be 4/9, which when added to the two cases will fetch the final answer.
Regards,
Harsha
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by hooliganpete » Fri Mar 19, 2010 1:28 pm
Thanks Harsha... as long as I'm getting the correct answer I suppose I can forgive myself for not paying close enough attention!
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