Inclusion-exclusion principle: Die thrown 10 times

Question

Question:
Use inclusion-exclusion to calculate the probability that the numbers 1,2 and 3 each appear at least once when a die is thrown 10 times.

My attempt:
$(6^{10}-3.5^{10} +3.4^{10}-3^{10})/6^{10} $

Is this correct?

Answer 1

Yes.

Let $E_i$ denote the event that number $i$ will not appear.

To be found is: $$1-\mathsf P(E_1\cup E_2\cup E_3)$$

With inclusion/exclusion and symmetry we find:$$\mathsf P(E_1\cup E_2\cup E_3)=3\mathsf P(E_1)-3\mathsf P(E_1\cap E_2)+\mathsf P(E_1\cap E_2\cap E_3)=$$$$3\cdot\left(\frac56\right)^{10}-3\cdot\left(\frac46\right)^{10}+\left(\frac36\right)^{10}$$

This agrees with your result.

Answer 2

Alternatively: $$P=1-[P(1')+P(2')+P(3')]+[P(1'\cap 2')+P(1'\cap 3')+P(2'\cap 3')]-$$ $$P(1'\cap 2'\cap 3')=1-3\left(\frac{5}{6}\right)^{10}+3\left(\frac{4}{6}\right)^{10}-\left(\frac{3}{6}\right)^{10}.$$