## Mitul 3 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...

1. Yahoo!

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2. Mitul

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

3. Yahoo!

i think the answer will be 0

4. Mitul

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

5. Yahoo!

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

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

7. Mitul

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

10. lgbasallote

uhhh this is just one event right?

11. Mitul

yeah...event that neither wins...

12. lgbasallote

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

13. lgbasallote

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

14. lgbasallote

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

16. apple_pi

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

17. lgbasallote

18. lgbasallote

what you did is probability that NOT both will win

19. waterineyes

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

oh im wrong again @lgbasallote :'(

21. Mitul

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

22. lgbasallote

there must be a silly trick to this :/

23. waterineyes

Subtract them from 1 each and then add them..

24. lgbasallote

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

25. lgbasallote

i really dont think this is one event

26. lgbasallote

could you post the whole question?/

27. Mitul

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

28. lgbasallote

hmm i knew it =_=

29. Mitul

this is the whole question @lgbasallote

30. apple_pi

um, tried my solution?

31. lgbasallote

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

32. Mitul

33. lgbasallote

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

34. waterineyes

$P(nor\; A) = \frac{1}{5}$ $P(nor \; B) = \frac{1}{6}$ $P(Neither \; A \; nor \;B) = P(nor \; A) + P(nor \; B)$

35. apple_pi

THINK SIMPLE! PROBABILITY TREE!

36. 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

37. lgbasallote

uhh should it be +

38. waterineyes

Sorry there will come multiplication..

39. Mitul

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

40. lgbasallote

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

41. lgbasallote

no paraphrasing...the exact one...

42. waterineyes

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

43. Mitul

@lgbasallote my qestion is perfect....the above is the solution which i gave for two cases....we need to consider both the probabilities

44. lgbasallote

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

45. Mitul

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]

46. apple_pi

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

47. telliott99

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

48. apple_pi

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