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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 = ?

Mathematics
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Please stop spamming and wait patiently for a tutor to assist you with your problems.
KKKKKKK sorry
Hey!! Xcuseme Who are you polpak ? May I know please?

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Other answers:

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