Probability of $12$ different dices containing at least one $5$ and at least one $4$?

Question

A set of $12 $ different standard dices ($6$ sufaces with numbers $1$ to $6$) is thrown. What is the probability if it contains at least one $5$ and at least one $4$?

My think that I have to choose the $2$ already known numbers $5$ and $4$ and multiply it by the rest of the possible results $6^{10}$. Finally dividing by all possibilities gives:

$\frac{\binom{12}{2} \cdot6^{10}}{6^{12}}= 1.833333$

which does not make sense. The correct answer should be $0.7834$.

I would like to understand the way how I can solve that task with combinatorics. I also appreciate other solutions.

Answer 1

You have to do some inclusion/exclusion.

Count all possibilities, subtract options without $4$, subtract options without $5$.

Inclusion/exclusion now says to ADD without $4$ and $5$ because you subtracted this twice instead of once.

$(6^{12} - 2*5^{12} + 4^{12})/6^{12} = 0.78339403706$

Answer 2

Hint -

Easy way to think is -

1 - (Any number on 12 dice except 5 + Any number on 12 dice except 4 + Any number on 12 dice except 4,5)

Answer 3

Let $A$ be the event "at least one $4$ is thrown" and $B$ be the event "at least one $5$ is thrown. We want to know $P(A \cap B)$. Also, for any event $E$, let $C(E)$ be the complement of that event.

Using inclusion-exclusion principle and some facts about complements, we have: $$P(A \cap B) = P(A)+P(B)-P(A \cup B) $$$$= 1-P(C(A))+1-P(C(B))-1+P(C(A\cup B))$$$$ = 1-P(C(A))-P(C(B))+P(C(A) \cap C(B)).$$

The probabilities in the last equation should be much easier to calculate, e.g., $C(A)$ is the event that no $4$'s are thrown.