anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1344504997861:dw|
anonymous
  • anonymous
not possible @Yahoo! 1-4/5-5/6 is negative...
anonymous
  • anonymous
i think the answer will be 0

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More answers

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

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