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anonymous
 4 years ago
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...
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1344504997861:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not possible @Yahoo! 14/55/6 is negative...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think the answer will be 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no cannot be....the event has occured....its not an impossible event...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u see there are only two chances either A will win or B will

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that can be one case but i nvr said that only A and B are playing.....

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhhh this is just one event right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah...event that neither wins...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yahoo's solution should've been right...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0also this doesnt make sense... 4/5 + 5/6 = 49/30

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the total probability should be less or equal to 1

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0draw a probability tree. Remember as we go along we multiply the values

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what you did is probability that NOT both will win

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[P(A \; and \;B \;win) = P(A) \times P(B)\] subtract it from 1 I think..

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@waterineyes u just did what @rsadhvika did...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there must be a silly trick to this :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Subtract them from 1 each and then add them..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we're thinking too complex...we need to open our brains to the possibilities :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i really dont think this is one event

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0could you post the whole question?/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yup @lgbasallote there are more than one case to this problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is the whole question @lgbasallote

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0um, tried my solution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the probabilities dont add up to less than or equal to 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think what @rsadhvika first posted should work then...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[P(nor\; A) = \frac{1}{5}\] \[P(nor \; B) = \frac{1}{6}\] \[P(Neither \; A \; nor \;B) = P(nor \; A) + P(nor \; B)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0THINK SIMPLE! PROBABILITY TREE!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Case 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry there will come multiplication..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhh can you post the WHOLE question with all these cases?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no paraphrasing...the exact one...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think we are making it complicated.. There is one and one case only..

anonymous
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you're suddenly giving situations...so i think what you originally posted isnt the full question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Case 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]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.04/5 > 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
 4 years ago
Best ResponseYou've already chosen the best response.0The example violates basic probability rules. If only one can win then P(A) + P(B) <= 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@telliott99, remember there are others in this 'race' or whatever it is.
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