vishweshshrimali5
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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polpak
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Please stop spamming and wait patiently for a tutor to assist you with your problems.
vishweshshrimali5
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KKKKKKK sorry
vishweshshrimali5
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Hey!! Xcuseme Who are you polpak ? May I know please?
polpak
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Just a lego guy
vishweshshrimali5
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What is lego guy ?
vishweshshrimali5
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I mean what does it mean ?
vishweshshrimali5
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Hey You have 580 medals !!!!!!!!!! wow
anonymous
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this question is a poser
polpak
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Others have more.
vishweshshrimali5
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Hello satellite please help me !
vishweshshrimali5
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What is a poser !
amingad
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are u studying iit coaching
vishweshshrimali5
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No but preparing for that but how do u know that man ?
anonymous
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if the first roll is a1, then probability is 1
if the first roll is a 2 then probability is
\[(\frac{5}{6})^3\]
amingad
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that type of question comes in iit
vishweshshrimali5
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Yaa u are right amingad !
amingad
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it is easy take 4 turn as 36c4
vishweshshrimali5
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How do u got that satellite please explain !
anonymous
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oh wait that is wrong. i need to read more carefully
vishweshshrimali5
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Please take you time
vishweshshrimali5
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How do you got that amingad
anonymous
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looks like coupon collectors problem
anonymous
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what is probability of a record. gives harmonic series
anonymous
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yikes now i have to think
vishweshshrimali5
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Amingad please explain step by step I can't understand it so fastly
vishweshshrimali5
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I am a neophyte in this field you may think that
amingad
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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
anonymous
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you lost me here for sure
vishweshshrimali5
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What do you mean to say satellite ?
vishweshshrimali5
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Okkkkk amingad now is there any step left ?
anonymous
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i mean i am lost completely.
vishweshshrimali5
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Ohhhh you mean that it is a nice question then thanks a lot
vishweshshrimali5
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Well satellite you can have a look at other questions
amingad
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12c3< m/ n so n has to be highest dice to be rolled
vishweshshrimali5
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Amingad will you please do me a favour
amingad
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yep it is tricky but good
vishweshshrimali5
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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
vishweshshrimali5
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Will you please do so ?
vishweshshrimali5
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Are you answering the question amingad
Zarkon
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\[\frac{7}{72}\]
amingad
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yep correct answer zarkon
Zarkon
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i know
Zarkon
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Just for fun...if we used an n-sided die the probability is
\[\frac{(n+1)(n+2)(n+3)}{24n^3}\]
Zarkon
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and if we rolled an n-sided die k times it would be
\[\frac{n(n+1)(n+2)\cdots(n+k-1)}{n^{k}k!}\]