Question

Brian rolls a number cube twice. What is the probability that the sum of the 2 rolls is less than 6 given that the first roll is a 3?

Answer 1

If the first roll is a three, then the only chance of the sum of the two numbers being less than six is if you roll a 1 or a 2.
That is 2/6 chance, or simplified, a 1/3 chance.

I hope this Helps!

Answer 2

Answer:  The required probability is $\dfrac{1}{3}.$

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

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{2}{6}=\dfrac{1}{3}.$

Thus, the required probability is $\dfrac{1}{3}.$