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Mitul
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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...
 2 years ago
 2 years ago
Mitul Group Title
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...
 2 years ago
 2 years ago

This Question is Closed

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
dw:1344504997861:dw
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
not possible @Yahoo! 14/55/6 is negative...
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
i think the answer will be 0
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
no cannot be....the event has occured....its not an impossible event...
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
u see there are only two chances either A will win or B will
 2 years ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
(14/5)(15/6)
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
that can be one case but i nvr said that only A and B are playing.....
 2 years ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
wait these are not mutually exclusive events
 2 years ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
oh ok gotcha
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
uhhh this is just one event right?
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
yeah...event that neither wins...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
yahoo's solution should've been right...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
also this doesnt make sense... 4/5 + 5/6 = 49/30
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
the total probability should be less or equal to 1
 2 years ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
P(both winning) = 4/5 * 5/6 P(both not winning) = 1 4/5 * 5/6
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
draw a probability tree. Remember as we go along we multiply the values
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
uhh no @rsadhvika..
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
what you did is probability that NOT both will win
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
\[P(A \; and \;B \;win) = P(A) \times P(B)\] subtract it from 1 I think..
 2 years ago

rsadhvika Group TitleBest ResponseYou've already chosen the best response.1
oh im wrong again @lgbasallote :'(
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
@waterineyes u just did what @rsadhvika did...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
there must be a silly trick to this :/
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Subtract them from 1 each and then add them..
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
we're thinking too complex...we need to open our brains to the possibilities :/
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
i really dont think this is one event
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
could you post the whole question?/
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
yup @lgbasallote there are more than one case to this problem
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
hmm i knew it =_=
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
this is the whole question @lgbasallote
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
um, tried my solution?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
the probabilities dont add up to less than or equal to 1
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
i got the answer....
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
i think what @rsadhvika first posted should work then...
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
\[P(nor\; A) = \frac{1}{5}\] \[P(nor \; B) = \frac{1}{6}\] \[P(Neither \; A \; nor \;B) = P(nor \; A) + P(nor \; B)\]
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
THINK SIMPLE! PROBABILITY TREE!
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
uhh should it be +
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Sorry there will come multiplication..
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
uhh can you post the WHOLE question with all these cases?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
no paraphrasing...the exact one...
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
I think we are making it complicated.. There is one and one case only..
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
@lgbasallote my qestion is perfect....the above is the solution which i gave for two cases....we need to consider both the probabilities
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.1
you're suddenly giving situations...so i think what you originally posted isnt the full question
 2 years ago

Mitul Group TitleBest ResponseYou've already chosen the best response.0
Case 1 1P(A)P(B) Case2...possibility that there are other playersss p(A and B cannot winwin)=1[4/5+5/64/5*5/6]
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

telliott99 Group TitleBest ResponseYou've already chosen the best response.0
The example violates basic probability rules. If only one can win then P(A) + P(B) <= 1
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.1
@telliott99, remember there are others in this 'race' or whatever it is.
 2 years ago
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