The probability of not acquiring a 5 is $(frac56)^3$, and I number the probcapability of acquiring at least one 5 is $1-(frac56)^3$, but I don"t understand exactly how to number out if it is rolled at leastern twice. Thoughts? Thanks in advance!


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The long winded brute pressure approach would be to add up the probabilities of the 4 outcomes that would give you the wanted result. $$5,5,5\5,5,x\5,x,5\x,5,5\ extWright here x is any number other than 5.$$

So, these outcomes would certainly have actually the probabilities of $$frac16^3\frac16cdotfrac16cdotfrac56\frac16cdotfrac56cdotfrac16\frac56cdotfrac16cdotfrac16$$

And you would add up these probabilities.

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The even more general means to look at it would be to, look into the binomial distribution, as said by a various answer to this question. Also,
robjohn offers a good explacountry on exactly how to find the probabilities for three and two 5"s.


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edited Jan 20 "15 at 2:34
answered Jan 20 "15 at 2:28
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turkeyhundtturkeyhundt
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The variety of ways to arrange $k$ fives out of $3$ dice is $displaystyleinom3k$

The probability of each plan, $k$ fives and $3-k$ others, is $displaystyleleft(frac16 ight)^kleft(frac56 ight)^3-k$

Hence, the probcapacity of gaining precisely $k$ fives out of $3$ dice is $displaystyleinom3kleft(frac16 ight)^kleft(frac56 ight)^3-k$

So the probcapability of gaining at leastern $2$ fives would be$$sum_k=2^3inom3kleft(frac16 ight)^kleft(frac56 ight)^3-k$$


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edited Jan 20 "15 at 9:43
answered Jan 20 "15 at 2:32
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robjohn♦robjohn
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There"s a famous circulation that answers the exact question you pose; namely the binomial circulation.


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edited Jan 20 "15 at 2:26
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Ahaan S. Rungta
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answered Jan 20 "15 at 2:22
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LeeLee
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Because you"re perdeveloping multiple experiments each with precisely 2 outcomes (the dice is a 5 vs. the dice is not a five), this is a timeless example of a Bernoulli trial.

The standard formula to calculate the probcapability is then to usage (from the Binomial distribution):

$$P\textthat A occurs precisely k times in n trials = left( eginarray*20c n \ k \ endarray ight) p^k (1 - p)^n-k$$

wbelow occasion $A$ is the trial and $p$ is the probability for the success of a solitary trial ($frac16$).

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In your situation, you desire to know the probcapability of it happening at least twice, so you must sum the chances for it happening precisely twice and specifically three times:

$$P\textthat A occurs specifically 2 times in 3 trials + P\textthat A occurs specifically 3 times in 3 trials$$

Therefore,$$eginalignP&= left( eginarray*20c 3 \ 2 \ endarray ight) p^2 (1 - p)^(3-2) + left( eginarray*20c 3 \ 3 \ endarray ight) p^3 (1 - p)^(3-3) \&= left( eginarray*20c 3 \ 2 \ endarray ight) cdot frac16^2 cdot frac56^1 + left( eginarray*20c 3 \ 3 \ endarray ight) cdot frac16^3 cdot frac56^0\&= 3 cdot frac136 cdot frac56 + 1 cdot frac1216 cdot 1\&= frac15216 + frac1216\&= frac16216\&= frac227 = 0,074... = 7,41 \% \endalign$$

As you deserve to see, this approach of calculation can unfortunately become very tedious for bigger examples, but it has actually the benefit of being a continuous strategy for any type of difficulties neighboring Bernoulli trials. The intuitive approaches pointed out in some of the other answers have the right to be faster for little situations wbelow all the feasible outcomes can be manually considered.