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9

The only way to get $k$ as an answer is if all rolls are $1$, so the probability is $\frac{1}{6^k}$. To get $k+1$ all rolls must be $1$ except for one roll which is a $2$; there are therefore $k$ different ways to do this and the probability is $\frac{k}{6^k}$.

7

A pentagonal bipyramid would work fine. The problem is that reading the result would be difficult. Dice roll on a surface and land on one of the faces, then you read the result (usually) on the face on top of the die (the standard d4 would be an exception). If you make a pentagonal bipyramid and it lands on one of the faces, however, the top of the die is ...

2

If you're looking for the number of events (when you roll, say $n$ dice) where a specific number, say $5$ must occur, you can just calculate the total number of events and remove the cases where $5$ does not occur. The total number of events is given by $6^n$ and the number of events where $5$ does not occur is given by $5^n$ (each number can be 1,2,3,4 or 6)...

2

If you want to use probability generating functions, then a single die corresponds to $\frac16(x+x^2+x^3+x^4+x^5+x^6)$ and so for $k$ dice: $$\left(\frac{x+x^2+x^3+x^4+x^5+x^6}{6}\right)^k=\left(\frac{x(1-x^6)}{6(1-x)}\right)^k$$ If you expand this you get \tfrac{1}{6^k}x^k +\tfrac{k}{6^k}x^{k+1}+\tfrac{k(k+1)}{2\times 6^k}x^{k+2}+\cdots + \tfrac{k(k+1)}{...

2

What the question is means that the die has 1 face which is 1 2 faces which are 2 3 faces which are 3 4 faces which are 4 for a total of 10 faces. Hope this clears it up.

1

The column with expected values provide the correct values. However, I would suggest simplifying your argument in the following way. \begin{equation} P_d(R) = P_d(\leq R)-P_d(\leq R-1) = \Bigl(\frac{R}{10}\Bigr)^{d} - \Bigl(\frac{R-1}{10}\Bigr)^{d} \end{equation} This simplifies calculating the expected value. \begin{equation} \mathop{\mathbb{...

1

Hint/Setup: This is an example of a question where it is far easier to calculate the probability of the complementary event, in other words the probability that this doesn't happen. If we didn't successfully get at least one $4$ and at least one $2$, then that means in the four throws of the dice, we didn't get any $4$'s or we didn't get any $2$'s, ...

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