1
$\begingroup$

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

Improve this question
$\endgroup$
2
  • $\begingroup$ would E in this case be $A \cap B$? $\endgroup$
    – user_1512314
    Commented Oct 11, 2019 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
    Commented Oct 11, 2019 at 7:40

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thanks for the help! is there a way to solve it if i dont treat the die's as fair? $\endgroup$
    – user_1512314
    Commented Oct 11, 2019 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 ♦
    Commented Oct 11, 2019 at 15:05
  • $\begingroup$ Thank you for your input! I thought as much that would be the case $\endgroup$
    – user_1512314
    Commented Oct 11, 2019 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
    Commented Oct 11, 2019 at 16:27
  • 1
    $\begingroup$ @user_1512314 if this answer solved your problem - or showed that it is not solvable - please accept the answer, so others will know that the problem is closed. $\endgroup$
    – Itamar Mushkin
    Commented Mar 26, 2020 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.