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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

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

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  • $\begingroup$ Thanks for the help! is there a way to solve it if i dont treat the die's as fair? $\endgroup$ – user_1512314 Oct 11 '19 at 13:17
  • $\begingroup$ +1. Alternatively you could state--correctly--that the answer is zero. Or one. Or any value in between. @user_1512314: dropping the fairness assumption, the problem is not solvable with the information given because you need to know (a) the probabilities for each die independently and (b) exactly how their probabilities depend on each other. There are two many possibilities to permit a unique numerical answer. $\endgroup$ – whuber Oct 11 '19 at 15:05
  • $\begingroup$ Thank you for your input! I thought as much that would be the case $\endgroup$ – user_1512314 Oct 11 '19 at 15:33
  • $\begingroup$ @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. $\endgroup$ – gunes Oct 11 '19 at 16:27

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