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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...

Mathematics
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not possible @Yahoo! 1-4/5-5/6 is negative...
i think the answer will be 0

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Other answers:

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

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