Question

A die is tossed and the number of dots facing up is noted.

a) Find the probability of the elementary events under theassumption that all faces of the die are equally likely to befacing up after a toss.
b) Find the probability of the elementary events under theassumption that the face with a single dot is twice as likely to befacing up as the rest of the faces.
c). Find the probability that the outcome of a toss is even underthe assumptions in parts a) and b) of this problem.

Answer 1

a) The probability of each elementary event when all faces are equally likely is 1/6. b) Under the assumption that the face with a single dot is twice as likely, the probabilities of the elementary events are 2/7 and 1/7. c) The probability of getting an even outcome is 1/2 in part a) and 2/7 in part b).

Step-by-step explanation:

a) In this case, since all faces of the die are equally likely, the probability of each elementary event is 1 divided by the total number of outcomes. Since the die has 6 faces, the total number of outcomes is 6. Therefore, the probability of each elementary event is 1/6.

b) Under the assumption that the face with a single dot is twice as likely to be facing up, the probability of each elementary event is not equal. The face with a single dot has a probability of 2/7, and the rest of the faces each have a probability of 1/7.

c) To find the probability that the outcome of a toss is even, we need to consider the outcomes that represent even numbers. In part a) where all faces are equally likely, the even outcomes are 2, 4, and 6. Therefore, the probability is 3/6 or 1/2. In part b) where the face with a single dot is twice as likely, the even outcomes are 2 and 4. Therefore, the probability is 2/7.