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

- jamiebookeater

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- anonymous

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- anonymous

not possible @Yahoo! 1-4/5-5/6 is negative...

- anonymous

i think the answer will be 0

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## More answers

- anonymous

no cannot be....the event has occured....its not an impossible event...

- anonymous

u see there are only two chances either A will win or B will

- rsadhvika

(1-4/5)(1-5/6)

- anonymous

that can be one case but i nvr said that only A and B are playing.....

- rsadhvika

wait these are not mutually exclusive events

- rsadhvika

oh ok gotcha

- lgbasallote

uhhh this is just one event right?

- anonymous

yeah...event that neither wins...

- lgbasallote

yahoo's solution should've been right...

- lgbasallote

also this doesnt make sense... 4/5 + 5/6 = 49/30

- lgbasallote

the total probability should be less or equal to 1

- rsadhvika

P(both winning) = 4/5 * 5/6
P(both not winning) = 1- 4/5 * 5/6

- anonymous

draw a probability tree. Remember as we go along we multiply the values

##### 1 Attachment

- lgbasallote

uhh no @rsadhvika..

- lgbasallote

what you did is probability that NOT both will win

- anonymous

\[P(A \; and \;B \;win) = P(A) \times P(B)\]
subtract it from 1 I think..

- rsadhvika

oh im wrong again @lgbasallote :'(

- anonymous

@waterineyes u just did what @rsadhvika did...

- lgbasallote

there must be a silly trick to this :/

- anonymous

Subtract them from 1 each and then add them..

- lgbasallote

we're thinking too complex...we need to open our brains to the possibilities :/

- lgbasallote

i really dont think this is one event

- lgbasallote

could you post the whole question?/

- anonymous

yup @lgbasallote there are more than one case to this problem

- lgbasallote

hmm i knew it =_=

- anonymous

this is the whole question @lgbasallote

- anonymous

um, tried my solution?

- lgbasallote

the probabilities dont add up to less than or equal to 1

- anonymous

i got the answer....

- lgbasallote

i think what @rsadhvika first posted should work then...

- 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

THINK SIMPLE! PROBABILITY TREE!

- 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

uhh should it be +

- anonymous

Sorry there will come multiplication..

- 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

uhh can you post the WHOLE question with all these cases?

- lgbasallote

no paraphrasing...the exact one...

- anonymous

I think we are making it complicated..
There is one and one case only..

- 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

you're suddenly giving situations...so i think what you originally posted isnt the full question

- 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

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

##### 1 Attachment

- anonymous

The example violates basic probability rules. If only one can win then
P(A) + P(B) <= 1

- anonymous

@telliott99, remember there are others in this 'race' or whatever it is.

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