Mitul
Probability of A winning the race is 4/5 and probabilty of B winning the race is 5/6. find the probability that neither wins...
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Yahoo!
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|dw:1344504997861:dw|
Mitul
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not possible @Yahoo! 1-4/5-5/6 is negative...
Yahoo!
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i think the answer will be 0
Mitul
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no cannot be....the event has occured....its not an impossible event...
Yahoo!
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u see there are only two chances either A will win or B will
rsadhvika
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(1-4/5)(1-5/6)
Mitul
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that can be one case but i nvr said that only A and B are playing.....
rsadhvika
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wait these are not mutually exclusive events
rsadhvika
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oh ok gotcha
lgbasallote
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uhhh this is just one event right?
Mitul
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yeah...event that neither wins...
lgbasallote
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yahoo's solution should've been right...
lgbasallote
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also this doesnt make sense... 4/5 + 5/6 = 49/30
lgbasallote
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the total probability should be less or equal to 1
rsadhvika
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P(both winning) = 4/5 * 5/6
P(both not winning) = 1- 4/5 * 5/6
apple_pi
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draw a probability tree. Remember as we go along we multiply the values
lgbasallote
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uhh no @rsadhvika..
lgbasallote
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what you did is probability that NOT both will win
waterineyes
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\[P(A \; and \;B \;win) = P(A) \times P(B)\]
subtract it from 1 I think..
rsadhvika
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oh im wrong again @lgbasallote :'(
Mitul
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@waterineyes u just did what @rsadhvika did...
lgbasallote
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there must be a silly trick to this :/
waterineyes
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Subtract them from 1 each and then add them..
lgbasallote
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we're thinking too complex...we need to open our brains to the possibilities :/
lgbasallote
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i really dont think this is one event
lgbasallote
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could you post the whole question?/
Mitul
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yup @lgbasallote there are more than one case to this problem
lgbasallote
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hmm i knew it =_=
Mitul
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this is the whole question @lgbasallote
apple_pi
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um, tried my solution?
lgbasallote
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the probabilities dont add up to less than or equal to 1
Mitul
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i got the answer....
lgbasallote
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i think what @rsadhvika first posted should work then...
waterineyes
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\[P(nor\; A) = \frac{1}{5}\]
\[P(nor \; B) = \frac{1}{6}\]
\[P(Neither \; A \; nor \;B) = P(nor \; A) + P(nor \; B)\]
apple_pi
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THINK SIMPLE! PROBABILITY TREE!
lgbasallote
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Case I
Probability A wont win is 1 - 4/5 = 1/5
Case II
probability B wont win is 1 - 5/6 = 1/6
so the probability both wont win is 1/6 x 1/5 = 1/30
lgbasallote
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uhh should it be +
waterineyes
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Sorry there will come multiplication..
Mitul
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this is not the question:
Solution:
the first case is when only A and B are playing
the second case is that there are other players
lgbasallote
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uhh can you post the WHOLE question with all these cases?
lgbasallote
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no paraphrasing...the exact one...
waterineyes
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I think we are making it complicated..
There is one and one case only..
Mitul
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@lgbasallote my qestion is perfect....the above is the solution which i gave for two cases....we need to consider both the probabilities
lgbasallote
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you're suddenly giving situations...so i think what you originally posted isnt the full question
Mitul
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Case -1
1-P(A)-P(B)
Case-2...possibility that there are other playersss
p(A and B cannot winwin)=1-[4/5+5/6-4/5*5/6]
apple_pi
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4/5 -> A will win (B cannot win)
1/5 -> A will not win (B still can)
5/6 -> B will win
1/6 -> B will not win (Neither A or B wins) [What you want]
If you take a look at my probability tree, you will see what I mean.
Remember you multiply the values as you go along.
So, YOUR ANSWER -> 1/6 * 1/5 = 1/30
telliott99
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The example violates basic probability rules. If only one can win then
P(A) + P(B) <= 1
apple_pi
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@telliott99, remember there are others in this 'race' or whatever it is.