# Probability of getting a 6 on at least one die from a pair of dependent dice

I am stuck on this particular question: Suppose you have two dice. These dice however are not independent: the probability that both dice will roll a 6 is 0.29. What is the probability that at least one of them rolls a 6 given that these dice are not independent? You can treat each die as fair when considering a single die's roll.

I was doing the following: Let $$A$$ be the event that the first die rolls a $$6$$ and let $$B$$ be the event that the second die rolls a $$6$$. Now, since $$P(A \cap B) = 0.29$$, I use the following to find when we get a 6 on the first die only:

$$P (A) = P(A \cap B) \ + P(A \cap B^c)$$

However, since we treat the roll of one die as being fair, $$P(A) = 1/6$$ which implies $$P(A \cap B^c)$$ is negative so I am definitely doing something wrong but I am not too sure what to do

• Maybe the $0.29$ is $P(A \mid B) = P(B \mid A)$? That is to say, maybe what is meant is that once you roll a $6$ on one die, the probability that the other die rolls a $6$ is $29/100$? It's hard to guess, since there's no physical model hinted at. Oct 11, 2019 at 5:15

Recall the Principle of Inclusion and Exclusion: $$\mathsf P(A\cup B)~{=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)\\=\tfrac {13}{300}}$$
So, no you are not doing anything wrong. You just cannot treat the die as having a fair marginal distribution yet a joint probability of $$0.29$$ .
• This doesn't really address OP's concern with negative probabilities. Notice that $29/100 > 1/6$. Oct 11, 2019 at 5:16