Answer: The required probability is [tex]\dfrac{1}{3}.[/tex]
Step-by-step explanation: Given that Brian rolls a number cube twice.
We are to find the probability that the sum of the 2 rolls is less than 6 given that the first roll is a 3.
Let, 'S' be the sample space for the experiment of rolling the second number cube when the first roll is a 3.
Then, S = {(3, 1), (3, 2), (3, 3), (3,4), (3, 5), (3, 6)} ⇒ n(S) = 6.
And, let 'E' be the event that the sum of the two rolls is less than 6.
Then, E = {(3, 1), (3, 2)} ⇒ n(E) = 2.
Therefore, the probability that the sum of the 2 rolls is less than 6 given that the first roll is a 3 is
[tex]P(E)=\dfrac{n(E)}{n(S)}=\dfrac{2}{6}=\dfrac{1}{3}.[/tex]
Thus, the required probability is [tex]\dfrac{1}{3}.[/tex]
