You roll a six sided die twice. What is the probability of rolling twice and getting a 1 or 2? What is the probability of not getting a 1 or 2? I've created a tree diagram and have come up with 20/36 chances of getting a 1 or 2. 16/36 getting anything else. Am I on the right track?

  • $\begingroup$ Your answers seem to be right $\endgroup$
    – ksha
    Apr 15, 2020 at 23:35

1 Answer 1


Yep, you'r right. You can use probability properties as well. Let

$A=$ the event that you get a $1$

$B=$ the event that you get a $2$

You can easily verify using a tree diagram that $P(A) =P(B) =\dfrac{11}{36}$, and $P(A\cap B) =\dfrac{2}{36}$. Now

$P(A\cup B) =P(A) +P(B) -P(A\cap B)$

$=\dfrac{11}{36} +\dfrac{11}{36} -\dfrac{2}{36} =\dfrac{20}{36}$.

And the probability of getting anything else, is the complement. That is

$P\left(\{ A \cup B\}^{c}\right) =1-P(A\cup B)$

$=1- \dfrac{20}{36} =\dfrac{16}{36}$.

  • $\begingroup$ Thank you so much! I thought it was too easy to be that. Felt like I was missing something $\endgroup$
    – user281310
    Apr 16, 2020 at 0:10

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