anonymous
  • anonymous
Roll a six-sided die 8 times. What is the probability that 6 will appear at least 5 times?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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DanJS
  • DanJS
five 6's in only 8 rolls is pretty difficult
anonymous
  • anonymous
indeed
DanJS
  • DanJS
the probability of rolling a certain number in a single role is 1 out of 6

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DanJS
  • DanJS
Say that happens on first role, the probability for the next roll is the same, 1/6 and for all 8 rolls the same
anonymous
  • anonymous
k
DanJS
  • DanJS
to get 5 of those you need that 1/6 to happen 5 times 1/6 * 1/6 * 1/6 * 1/6 * 1/6
anonymous
  • anonymous
1/7776
DanJS
  • DanJS
yeah,. that is for rolling all in a row...
DanJS
  • DanJS
the change of not getting 6 is 5/6 for each roll
anonymous
  • anonymous
ok
DanJS
  • DanJS
so above we considered 5 rolls , want 8 total rolls the 3 left can be misses with a 5/6 chance
DanJS
  • DanJS
so 5 times chance of getting 6 + 3 times chance of missing is getting 'at least' five sixes out of 8 rolls
DanJS
  • DanJS
[1/6 ]^5 * [5/6] ^3
anonymous
  • anonymous
in other word is 5(9)?
DanJS
  • DanJS
8 rolls total. multiply the probability for each one together. 5 of them should be 1/6 and 8 of them should be 5/6
anonymous
  • anonymous
i got like 125/1679616
anonymous
  • anonymous
ok
DanJS
  • DanJS
yeah , small chance
anonymous
  • anonymous
do that mean i need to add 5,6,7,and 8?
DanJS
  • DanJS
that fraction is about 0.00007
anonymous
  • anonymous
that number is too long
anonymous
  • anonymous
perhaps im wrong
DanJS
  • DanJS
were you shown the formula with the n and k , n rolls , k times something occurs
anonymous
  • anonymous
\[\left(\begin{matrix}n \\ k\end{matrix}\right) p^k(1-p)^{n-k}\] this formula ?
DanJS
  • DanJS
that looks good if you see, it is the same thing we did above
DanJS
  • DanJS
p = success chance, so (1-p) is the chance of fail k = # success n = # failed p=1/6 , (1-p) = 5/6 k= 5 rolls n = 3 rolls
anonymous
  • anonymous
ok got 0.00461
kropot72
  • kropot72
That is correct.
anonymous
  • anonymous
k thx so much
kropot72
  • kropot72
You're welcome :)

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