Calculate probability rolling a die $10$ times I you roll a die $10$ times:
What is the probability of getting at least a  one and at least a two?
I tried to use the complement rule and find the probabilities of not getting a one and not getting a two. As I have four numbers left the probability is 
$= 1  - \frac{4^{10}}{6^{10}} = 0.9826$ . But the right answers is $0.6943$
Thank you.
 A: Consider four possible outcomes of the $10$ rolls:


*

*You roll no ones and no twos. The probability of this is 
$p_0 = \frac{4^{10}}{6^{10}}$
as you correctly derived from the fact that the outcome of each roll
is three, four, five, or six.

*You roll at least one "one", but you do not roll any twos. Let this have probability $p_1$.

*You roll at least one "two", but you do not roll any ones. Let this have probability $p_2$.

*You roll at least one "one" and also at least one "two". Let this have probability $p_3$.
The last case is the one whose probability you want to measure,
$p_3 = 1 - (p_0 + p_1 + p_2).$
As you can see, $p_1 = p_2$. To calculate $p_1$, you can take the probability
that you do not roll any twos, and subtract the probability that you do not roll any ones or twos. That is, if $A_1$ is the event that at least one "one" is rolled,
and $A_1$ is the event that at least one "two" is rolled, then
$$P(A_2^\complement)
 = P(A_1 \cap A_2^\complement) + P(A_1^\complement \cap A_2^\complement)$$
and therefore
$$p_1 = P(A_1 \cap A_2^\complement) 
= P(A_2^\complement) - P(A_1^\complement \cap A_2^\complement).$$
If you write out $p_0 + p_1 + p_2$ in this manner, you will find that the sum
simplifies to
$$p_0 + p_1 + p_2 =
 P(A_2^\complement) + P(A_2^\complement) - P(A_1^\complement \cap A_2^\complement),$$
which is an application of the Inclusion-Exclusion principle to this problem.
A: $$\scriptsize\rm P(1.2)=P((1'+2')')=1-P(1'+2')=1-[P(1')+P(2')-P(1'.2')]=1-\left[\frac{5^{10}}{6^{10}}+\frac{5^{10}}{6^{10}}-\frac{4^{10}}{6^{10}}\right]=\frac{6^{10}-2\times5^{10}+4^{10}}{6^{10}}\\\approx \color{red}{0.6943}\color{grey}{30364...}$$
