# 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

• would E in this case be $A \cap B$? – user_1512314 Oct 11 '19 at 5:20
• Joke: $P(A \cup B) = P(A) + P(B) - P(AB) =\frac 16 + \frac 16 - .29 = \frac 13 - .29 = 0.04333333$ – BruceET Oct 11 '19 at 7:40

You are correct. $$P(A\cap B)\leq P(A)$$. Your way or another (as below), we can find an upper bound on the event that both dice will be $$6$$: $$P(A\cap B)=P(A)P(B|A)=(1/6)P(B|A)\leq 1/6$$ So the problem has contradictions.
• @user_1512314 knowing that they’re dependent is not sufficient, since we don’t know the dependence relation. Let’s say $B=A$, then the answer is $0.29$, (assuming not fair dice due to contradictory behavior stated in the answer). This is a simple case, and there are infinitely others when we don’t know the dependence relation. – gunes Oct 11 '19 at 16:27