Question

Tom rolls a number cube twice. What is the probability that the sum of the two rolls is less than 9, given that the first roll is a 6? •1/3 •2/3 •1/4 •1/6

Answer 1

2/6 that is 1/3 will be the answer

Answer 2

Answer: 1/3

Step-by-step explanation:

Since, when the two dice are rolled,

The total outcomes, n(S) = 6 × 6 = 36,

Now, the possible outcomes that the sum of the two rolls is less than 9 when two dice are rolled

E = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (5,1), (5,2), (5,3), (6,1), (6,2)}

Now, The possible outcomes that the first roll is a 6,

F = { (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

n(F) = 6,

⇒ $P(F)=\frac{n(F)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

Since, E ∩ F = { (6,1), (6,2) }

⇒ n( E ∩ F ) = 2

Thus,

$P(E\cap F) = \frac{n(E\cap F)}{n(S)}= \frac{2}{36}=\frac{1}{18}$

Hence, the probability that the sum of the two rolls is less than 9, given that the first roll is a 6

$P(\frac{E}{F})=\frac{P(E\cap F)}{P(F)}=\frac{1/18}{1/6}=\frac{1}{3}$