Probability of obtaining a double six in at least two throws The question: A pair of fair dice is thrown 10 times. What is the probability of obtaining a double six in at least two throws?  
My attempt: Let X denote the total number of double sixes obtained. On any throw, the probability of obtaining a double six is $\frac 1{36}$ and thus, the probability of not obtaining a double six is $\frac {35}{36}$.
$P(X=1) = 10(\frac 1{36})(\frac {35}{36})^9$ since the double six can be obtained in any of the 10 throws.
$P(X=0) = (\frac {35}{36})^{10}$
So $P(X \ge 2) = 1 - P(X=1) - P(X=0) = 1 - (\frac {10}{36})(\frac{35}{36})^9 - (\frac{35}{36})^{10}$  
However, the school's solution says that $P(X \ge 2) = 1 - (\frac {1}{36})(\frac{35}{36})^9 - (\frac{35}{36})^{10}$  
Which is correct? And why?
 A: Your answer is correct.   The school’s solution is in error.
$\begin{align}
\mathsf P(X\geq 2) & = 1 - \mathsf P(X\leq 1) & \text{By the Law of Complements}
\\ & = 1 - {10\choose 0}p^0(1-p)^{10} - {10\choose 1}p^1(1-p)^9 & \text{because it is a binomial distribution}
\\ & = 1- \frac{35^{10}}{36^{10}}-\frac{10\cdot 35^9}{36^{10}}
\end{align}$
A: The reason you have to do $$P(X \ge 2) = 1 - (\frac {10}{36})(\frac{35}{36})^9 - (\frac{35}{36})^{10}$$ Is because the of this: $$P(X\ge2)=1-P(X\le1)=1-(\frac {10}{36})(\frac{35}{36})^9 - (\frac{35}{36})^{10}$$ and the reason the $10$ is there is because of this: $${10\choose 1} \cdot{1\over 36}^1{35\over 36}^9$$ So your answer is the correct answer.
A: I think your answer is correct and the school's answer is a mistake.
Since you already correctly answered the question, if we have to write a lot more about why, by the school's method (neglecting the binomial coefficient),
$$
P(X>10) = 1 - \sum_{n=0}^{10}(1/36)^{n}(35/36)^{10-n} =1-(1/36)^{10}\sum_{n=0}^{10}35^n\neq 0
$$
so the school's method gives you a non-zero probability of an impossible event.
