Probabilities of no ace and exactly not one ace in throwing 10 dies

Question

The probabilities i am looking for are used for a problem in Feller's introduction to probability that asks for the probability $p$ of finding $2$ or more aces in throwing $10$ dies. For this problem i considered that:

$p = 1-x-y$

Where $x$ is the probability of finding no ace at all, and $y$ being the probability of finding exactly one ace. Of course $x = \frac{5^{10}}{6^{10}}$... i think. But i am a little more confused about finding $y$, more because the answer to this exercise is that: $p = 1- 10\times\frac{5^9}{6^{10}-5^{10}} $ but i don't really understand where this result can come from, more considering the result i got for $x$

Answer 1

The book is wrong. And your factorization is slightly off. (You need a plus sign inside the parenthesis).$1-\frac{5^{10}}{6^{10}}-10*\frac{5^9}{6^{10}}=1-\frac{5^9}{6^{10}}\left(5+10\right)\approx 0.5154833$ whereas the answer in the book is $1-10*\frac{5^9}{6^{10}-5^{10}}\approx 0.6147724$.

This can be computed also without using complementary events as ${10\choose2}(1/6)^2(5/6)^8+...+{10\choose 10}(1/6)^{10}(5/6)^0$. Using code, this is

> pbinom(1, 10, 1/6, lower.tail=FALSE)
[1] 0.5154833

Answer 2

Hint

If the probability of an event occurring is $p \in (0,1)$ and $q = (1-p)$ then
the probability of having exactly $k$ successes in $n$ trials : $k \in \{0,1,2,\cdots, n\}$ is

$$\binom{n}{k}p^k q^{(n-k)}.$$

Answer 3

Use $\begin{pmatrix}n \\ k\end{pmatrix}p^kq^{n-k}$ with n= 10, k= 1, p= 5/6 and q= 1/6.