4 years ago

If you roll a six-sided die, what is the probability that you will obtain a number greater than 2

Mathematics

1 answer:

Iteru [2.4K]4 years ago

6 0

thee are 6 numbers on a die

4 are greater than 2

so you would have a 4/6, reduced to 2/3 probability

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zalisa [80]

Perhaps you know that

$S_2 = \displaystyle\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6$

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Then the problem is trivial, since

$\displaystyle\sum_{k=1}^n k^2(k+1) = S_2 + S_3 \\\\ = \frac{2n(n+1)(2n+1)+3n^2(n+1)^2}{12} \\\\ = \frac{n(n+1)\big((2(2n+1)+3n(n+1)\big)}{12} \\\\ = \frac{n(n+1)\big(4n+2+3n^2+3n\big)}{12} \\\\ = \frac{n(n+1)(3n^2+7n+2)}{12} \\\\ = \frac{n(n+1)(3n+1)(n+2)}{12}$

Then

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3 years ago