Question

We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely. (a) Given that the roll results in a sum of 3 or more, find the conditional probability that doubles (the first and the second rolls result in the same number) are rolled. (b) Given that the two dice land on different numbers, find the conditional probability that the sum is 5.

Answer 1

Answer:

a) $\frac{1}{7}$

b) $\frac{2}{15}$

Step-by-step explanation:

Given : We roll two fair 6-sided dice. Each of the 36 possible outcomes is assumed to be equally likely.

The outcomes are :

(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) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

(a) Given that the roll results in a sum of 3 or more, find the conditional probability that doubles (the first and the second rolls result in the same number) are rolled.

Let P(A) be the probability of event that the roll results in a sum of 3 or more.

Except (1,1) rest sum is 3 or greater than 3.

So, $P(A)=\frac{35}{36}$

Now, P(A and B) is that the roll results in a sum of 3 or more and that doubles (the first and the second rolls result in the same number) are rolled.

i.e. (2,2), (3,3), (4,4), (5,5), (6,6) - 5

$P(A\cap B)=\frac{5}{36}$

The conditional probability is given by,

$P(B|A)=\frac{P(A\cap B)}{P(A)}$

$P(B|A)=\frac{\frac{5}{36}}{\frac{35}{36}}$

$P(B|A)=\frac{5}{35}$

$P(B|A)=\frac{1}{7}$

(b) Given that the two dice land on different numbers, find the conditional probability that the sum is 5.

Let P(A) be the probability of event that two dice land on different numbers.

Except (1,1),(2,2), (3,3), (4,4), (5,5), (6,6) rest two dice land on different numbers.

So, $P(A)=\frac{30}{36}$

Now, P(A and B) is that two dice land on different numbers and the sum is 5.

i.e. (1,4), (2,3), (3,2), (4,1) - 4

$P(A\cap B)=\frac{4}{36}$

The conditional probability is given by,

$P(B|A)=\frac{P(A\cap B)}{P(A)}$

$P(B|A)=\frac{\frac{4}{36}}{\frac{30}{36}}$

$P(B|A)=\frac{4}{30}$

$P(B|A)=\frac{2}{15}$