Question

We roll a fair die four times.

(a) Describe the sample space Ohm and the probability measure P that models this experiment (i.e., give the value P(omega) for each outcome omega elementof Ohm).

(b) Let A be the event that there are at least two fives among the four rolls. Let B be the event that there is at most one five among the four rolls. Find the probabilities P(A) and P(B), by finding the ratio of the number of favorable outcomes to the total, as in the Equally Likely Outcomes rule.

(c) What is the set AUB? What equality should P(A) and P(B) satisfy? Check that your answers to part (b) satisfy this equality.

Answer 1

The sample space Ohm for rolling a fair die four times is described as Ohm = {1,2,3,4,5,6} x {1,2,3,4,5,6} x {1,2,3,4,5,6} x {1,2,3,4,5,6}. The probability measure P assigns a probability of 1/1296 to each outcome. The probabilities of events A and B are 1/216 and 5/216, respectively.

Step-by-step explanation:

a) The sample space, Ohm, for rolling a fair die four times can be represented as Ohm = {1,2,3,4,5,6} x {1,2,3,4,5,6} x {1,2,3,4,5,6} x {1,2,3,4,5,6} as there are 6 possible outcomes for each roll. The total number of outcomes in Ohm is 6^4 = 1296. The probability measure, P, is the function that assigns a probability to each outcome. Since the die is fair, each outcome is equally likely, so P(omega) = 1/1296 for each outcome omega in Ohm.

b) Let A be the event that there are at least two fives among the four rolls. To find P(A), we need to find the number of favorable outcomes and divide it by the total number of outcomes. The number of favorable outcomes is the number of ways we can choose at least two fives from the four rolls, which is the combination C(4,2) = 6. The total number of outcomes is 6⁴ = 1296. Therefore, P(A) = 6/1296 = 1/216.

Let B be the event that there is at most one five among the four rolls. To find P(B), we need to find the number of favorable outcomes and divide it by the total number of outcomes. The number of favorable outcomes is the number of ways we can choose at most one five from the four rolls, which is the combination C(4,0) + C(4,1) = 1 + 4 = 5. The total number of outcomes is 6^4 = 1296. Therefore, P(B) = 5/1296 = 5/216.

Answer 2

A and B are complementary events

P(A)+P(B) =1

Step-by-step explanation:

a) sample space will consist of 6x6x6x6 events as

(1,1,1,1) to (6,6,6,6)

P(each event) =
`(1)/(6^4)`

b) A- there are atleast two 5 among the four rolls.

B = atmost one 5 among four rolls

Let X be the no of 5's in the four rolls. X is binomial since each die is independent of the other with p = 1/6 and n =4

P(A) =
`P(X\geq 2)\\=\Sigma _2^5 (5Cr)(1)/(6) ((5)/(6) )^(4-r)`

=0.131944

P(B) =
`P(X\leq 1) = 0.868094`

We find that A and B are complementary events.

c) P(AUB) =1 since A and B are complementary