Please I need help on how to go about solving this question:
A dice has $n$ sides with numbers $1,2,3,4,...,n$ marked on it. If the probability that a particular number appears at least once when the dice is rolled thrice is $217/729$, then how many sides does the dice have?
I was thinking that if the number was one, then this is how I can go about the solution: the probability of one appearing at least once will be something like this
$$\frac{1}{n}\times \frac{2}{n}\times \frac{3}{n}$$
Then I equate it to $217/729$ and find for $n$:
$$\frac{1}{n}\times \frac{2}{n}\times \frac{3}{n}=\frac{217}{729}$$
But since the question didn't give any number, but rather said "a particular number appears at least once" so I became a bit confused on how to solve it. Please any help on how to go about it?
$\endgroup$ 52 Answers
$\begingroup$We need to assume here that the dice is fair, that is, that when we throw it the probability that a particular side appear is the same for every side. This mean that the probability that, after a throw, the number that appear is one will be $1/n$, where $n$ is the number of sides of the dice.
Then the probability that a one appear at least once in three throws will be the complementary probability that no one appear at all, that is$$ 1-\left(\frac{n-1}{n}\right)^{3} $$
Equating the above with the given probability, we have that
$$ \frac{n-1}{n}=\sqrt[3]{1-\frac{217}{729}}\iff n=\frac1{1-\sqrt[3]{1-\frac{217}{729}}}=9 $$
$\endgroup$ $\begingroup$If you roll an $n$-sided die three times, the probability that a particular number (it doesn't matter which number that is, if the die is fair) doesn't appear at all is $(\frac{n-1}{n})^3$. The inverse of this, $1-(\frac{n-1}{n})^3$, is the probability that the given number appears at least once.
So, you want to find $n \in \mathbb{N}$ such that $1-(\frac{n-1}{n})^3 = \frac{217}{729}$. Multiplying both sides by $729n^3$ to get rid of the fractions, and doing some more algebra, gives the cubic equation $217 n^3 - 2187 n^2 + 2187 n - 729 = 0$. This has three roots, but two of them are complex and thus not a valid answer the question. The other one is $n = 9$.
If using the Cubic Formula or trying to factor the polynomial is too tedious for you, an easier approach is to note that $\sqrt[3]{729} = 9$, so the number of sides must be a multiple of 9. So try plugging in $n = 9, 18, 27, 36, ...$ until you get a solution.
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