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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...

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  1. Yahoo!
    • 3 years ago
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    |dw:1344504997861:dw|

  2. Mitul
    • 3 years ago
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    not possible @Yahoo! 1-4/5-5/6 is negative...

  3. Yahoo!
    • 3 years ago
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    i think the answer will be 0

  4. Mitul
    • 3 years ago
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    no cannot be....the event has occured....its not an impossible event...

  5. Yahoo!
    • 3 years ago
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    u see there are only two chances either A will win or B will

  6. rsadhvika
    • 3 years ago
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    (1-4/5)(1-5/6)

  7. Mitul
    • 3 years ago
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    that can be one case but i nvr said that only A and B are playing.....

  8. rsadhvika
    • 3 years ago
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    wait these are not mutually exclusive events

  9. rsadhvika
    • 3 years ago
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    oh ok gotcha

  10. lgbasallote
    • 3 years ago
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    uhhh this is just one event right?

  11. Mitul
    • 3 years ago
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    yeah...event that neither wins...

  12. lgbasallote
    • 3 years ago
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    yahoo's solution should've been right...

  13. lgbasallote
    • 3 years ago
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    also this doesnt make sense... 4/5 + 5/6 = 49/30

  14. lgbasallote
    • 3 years ago
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    the total probability should be less or equal to 1

  15. rsadhvika
    • 3 years ago
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    P(both winning) = 4/5 * 5/6 P(both not winning) = 1- 4/5 * 5/6

  16. apple_pi
    • 3 years ago
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    draw a probability tree. Remember as we go along we multiply the values

  17. lgbasallote
    • 3 years ago
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    uhh no @rsadhvika..

  18. lgbasallote
    • 3 years ago
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    what you did is probability that NOT both will win

  19. waterineyes
    • 3 years ago
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    \[P(A \; and \;B \;win) = P(A) \times P(B)\] subtract it from 1 I think..

  20. rsadhvika
    • 3 years ago
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    oh im wrong again @lgbasallote :'(

  21. Mitul
    • 3 years ago
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    @waterineyes u just did what @rsadhvika did...

  22. lgbasallote
    • 3 years ago
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    there must be a silly trick to this :/

  23. waterineyes
    • 3 years ago
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    Subtract them from 1 each and then add them..

  24. lgbasallote
    • 3 years ago
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    we're thinking too complex...we need to open our brains to the possibilities :/

  25. lgbasallote
    • 3 years ago
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    i really dont think this is one event

  26. lgbasallote
    • 3 years ago
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    could you post the whole question?/

  27. Mitul
    • 3 years ago
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    yup @lgbasallote there are more than one case to this problem

  28. lgbasallote
    • 3 years ago
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    hmm i knew it =_=

  29. Mitul
    • 3 years ago
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    this is the whole question @lgbasallote

  30. apple_pi
    • 3 years ago
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    um, tried my solution?

  31. lgbasallote
    • 3 years ago
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    the probabilities dont add up to less than or equal to 1

  32. Mitul
    • 3 years ago
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    i got the answer....

  33. lgbasallote
    • 3 years ago
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    i think what @rsadhvika first posted should work then...

  34. waterineyes
    • 3 years ago
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    \[P(nor\; A) = \frac{1}{5}\] \[P(nor \; B) = \frac{1}{6}\] \[P(Neither \; A \; nor \;B) = P(nor \; A) + P(nor \; B)\]

  35. apple_pi
    • 3 years ago
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    THINK SIMPLE! PROBABILITY TREE!

  36. lgbasallote
    • 3 years ago
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    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
    • 3 years ago
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    uhh should it be +

  38. waterineyes
    • 3 years ago
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    Sorry there will come multiplication..

  39. Mitul
    • 3 years ago
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    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
    • 3 years ago
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    uhh can you post the WHOLE question with all these cases?

  41. lgbasallote
    • 3 years ago
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    no paraphrasing...the exact one...

  42. waterineyes
    • 3 years ago
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    I think we are making it complicated.. There is one and one case only..

  43. Mitul
    • 3 years ago
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    @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
    • 3 years ago
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    you're suddenly giving situations...so i think what you originally posted isnt the full question

  45. Mitul
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    The example violates basic probability rules. If only one can win then P(A) + P(B) <= 1

  48. apple_pi
    • 3 years ago
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    @telliott99, remember there are others in this 'race' or whatever it is.

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