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# Maths (probability ) pls help

Thu 7 Jun, 2012 02:23 am
A problem was given to "A" whose probability to solve it is 2/3. If "A' couldnt solve it, the same is given to "B" whose probability to solve it when "A' could not solve is 5/7. If "B" couldnt solve it then, the same is given to "C" whose probability to solve it then is 1/10. what is the probability that problem gets solved.
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Type: Discussion • Score: 2 • Views: 1,679 • Replies: 6
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engineer

1
Thu 7 Jun, 2012 05:11 am
@sandy111,
The probability of it not being solved is 1/3 x 2/7 x 9/10 = 6/70 so the probability of solving it is 64/70.
sandy111

1
Thu 7 Jun, 2012 06:47 am
@engineer,
hello....
This is not correct. Your answer is correct if the situation is like a problem is given to 3 persons simultaneously and all are working it together. but here condition is different. If first student is unable to solve, it is given to second (first not involved) and if second is unable to solve, it is given to third (second not involved).....Plss...solve it..
maxdancona

1
Thu 7 Jun, 2012 07:06 am
@sandy111,
Look at engineer's solution again. It is correct (hint, he multiplied the numbers subracted from 1).
sandy111

1
Thu 7 Jun, 2012 07:26 am
@maxdancona,
Hii.
i understood how he solved but he understood question differently... Actual problem i already stated......Let me explain...In my problem, B solved only when A is unable to and that time A's role is over.. Same C solved wen B unable to solve. that time B role is over.... But engineer solution presumes that when A solves , B and C also try to solve but unable to solve. When B solves, A and C also try to solve, but unable to solve...............
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engineer

1
Thu 7 Jun, 2012 07:35 am
@sandy111,
If you are trying to compute the probability of not solving the problem, there is no difference between them working sequentially or in parallel - they all must still fail to solve the problem. That's why you approach problems like this from the fail perspective instead of the pass perspective. If you want to do it the much harder way, it looks like this:

P = A + A'B + A'B'C

P = 2/3 + (1/3)(5/7) + (1/3)(2/7)(1/10) = 64/70
sandy111

1
Thu 7 Jun, 2012 07:39 am
@engineer,
superb.........great...........
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