Question

Exercise 1.1. We roll a fair die twice. Describe a sample space ⌦ and a probability measure P to model this experiment. Let A be the event that the second roll is larger than the first. Find the probability P(A) that the event A occurs.

Answer 1

Answer:

$P=\frac{1}{36}$

$P(A)=\frac{5}{12}$

Step-by-step explanation:

The sample space of rolling a dice twice is:

S = {(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)}

Sample size, n = 36.

Each sample has an equal chance of being selected.

So the probability of selecting any of the 36 sample is, $P=\frac{1}{36}$.

Now an event A is defined as the second roll being larger than the first.

The sample space for A is:

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

The sample size is, n = 15.

The probability of the event A is:

$P(A)=\frac{15}{36}=\frac{5}{12}$