0
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – Brian Tung
    Commented Oct 11, 2019 at 5:15

1 Answer 1

1
$\begingroup$

Recall the Principle of Inclusion and Exclusion: $$\mathsf P(A\cup B)~{=\mathsf P(A)+\mathsf P(B)-\mathsf P(A\cap B)\\=\tfrac {13}{300}}$$

Notice the probability for the union is less than the probability for the intersection.   It should not be.

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 $0.29$ .

$\endgroup$
2
  • $\begingroup$ This doesn't really address OP's concern with negative probabilities. Notice that $29/100 > 1/6$. $\endgroup$
    – Brian Tung
    Commented Oct 11, 2019 at 5:16
  • $\begingroup$ Thanks for the help! Is there a way to solve it if i dont treat a single die as fair? $\endgroup$ Commented Oct 11, 2019 at 13:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .