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vishweshshrimali5

  • 4 years ago

A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceeding it is m/n in the lowest terms. Then n -10m = ?

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  1. polpak
    • 4 years ago
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    Please stop spamming and wait patiently for a tutor to assist you with your problems.

  2. vishweshshrimali5
    • 4 years ago
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    KKKKKKK sorry

  3. vishweshshrimali5
    • 4 years ago
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    Hey!! Xcuseme Who are you polpak ? May I know please?

  4. polpak
    • 4 years ago
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    Just a lego guy

  5. vishweshshrimali5
    • 4 years ago
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    What is lego guy ?

  6. vishweshshrimali5
    • 4 years ago
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    I mean what does it mean ?

  7. vishweshshrimali5
    • 4 years ago
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    Hey You have 580 medals !!!!!!!!!! wow

  8. satellite73
    • 4 years ago
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    this question is a poser

  9. polpak
    • 4 years ago
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    Others have more.

  10. vishweshshrimali5
    • 4 years ago
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    Hello satellite please help me !

  11. vishweshshrimali5
    • 4 years ago
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    What is a poser !

  12. amingad
    • 4 years ago
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    are u studying iit coaching

  13. vishweshshrimali5
    • 4 years ago
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    No but preparing for that but how do u know that man ?

  14. satellite73
    • 4 years ago
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    if the first roll is a1, then probability is 1 if the first roll is a 2 then probability is \[(\frac{5}{6})^3\]

  15. amingad
    • 4 years ago
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    that type of question comes in iit

  16. vishweshshrimali5
    • 4 years ago
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    Yaa u are right amingad !

  17. amingad
    • 4 years ago
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    it is easy take 4 turn as 36c4

  18. vishweshshrimali5
    • 4 years ago
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    How do u got that satellite please explain !

  19. satellite73
    • 4 years ago
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    oh wait that is wrong. i need to read more carefully

  20. vishweshshrimali5
    • 4 years ago
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    Please take you time

  21. vishweshshrimali5
    • 4 years ago
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    How do you got that amingad

  22. satellite73
    • 4 years ago
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    looks like coupon collectors problem

  23. vishweshshrimali5
    • 4 years ago
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    It may be

  24. satellite73
    • 4 years ago
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    what is probability of a record. gives harmonic series

  25. satellite73
    • 4 years ago
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    yikes now i have to think

  26. vishweshshrimali5
    • 4 years ago
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    Amingad please explain step by step I can't understand it so fastly

  27. vishweshshrimali5
    • 4 years ago
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    I am a neophyte in this field you may think that

  28. amingad
    • 4 years ago
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    fair dice consist of 6 numbers . so the the no of times get the highest probability of number will be 24 c4 my mistake i wrote 36 c4

  29. satellite73
    • 4 years ago
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    you lost me here for sure

  30. vishweshshrimali5
    • 4 years ago
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    What do you mean to say satellite ?

  31. vishweshshrimali5
    • 4 years ago
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    Okkkkk amingad now is there any step left ?

  32. satellite73
    • 4 years ago
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    i mean i am lost completely.

  33. vishweshshrimali5
    • 4 years ago
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    Ohhhh you mean that it is a nice question then thanks a lot

  34. vishweshshrimali5
    • 4 years ago
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    Well satellite you can have a look at other questions

  35. amingad
    • 4 years ago
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    12c3< m/ n so n has to be highest dice to be rolled

  36. vishweshshrimali5
    • 4 years ago
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    Amingad will you please do me a favour

  37. amingad
    • 4 years ago
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    yep it is tricky but good

  38. vishweshshrimali5
    • 4 years ago
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    can you write the answer in a single post and with each step explained Because my teacher says that you will have to answer the question with each step explained or you will get no marks

  39. vishweshshrimali5
    • 4 years ago
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    Will you please do so ?

  40. vishweshshrimali5
    • 4 years ago
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    Are you answering the question amingad

  41. Zarkon
    • 4 years ago
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    \[\frac{7}{72}\]

  42. amingad
    • 4 years ago
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    yep correct answer zarkon

  43. Zarkon
    • 4 years ago
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    i know

  44. Zarkon
    • 4 years ago
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    Just for fun...if we used an n-sided die the probability is \[\frac{(n+1)(n+2)(n+3)}{24n^3}\]

  45. Zarkon
    • 4 years ago
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    and if we rolled an n-sided die k times it would be \[\frac{n(n+1)(n+2)\cdots(n+k-1)}{n^{k}k!}\]

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