Three identical chips, each with 0 on one side and 1 on the other, are to be tossed simultaneously. A. writes the sample space. B. what is the probability of observing the sum of the face-up numbers to equal to 2? C. What is the probability of observing the sum of the face up numbers to be at least 1? d. what is the probability of observing the sum of the face up numbers equal to 2.5?
A. To write the sample space, we need to consider all the possible outcomes of tossing the three chips simultaneously. Each chip has two possible outcomes: either 0 or 1.
The sample space would be the set of all possible combinations of these outcomes. Since there are three chips, and each chip has two possible outcomes, the size of the sample space would be 2^3 (2 raised to the power of 3), which is 8.
The sample space can be written as follows:
{000, 001, 010, 011, 100, 101, 110, 111}
B. The event of observing the sum of the face-up numbers to equal 2 can occur in the following ways:
(1, 1, 0), (1, 0, 1), and (0, 1, 1).
Thus, there are 3 favorable outcomes. Since the sample space has 8 possible outcomes, the probability of observing a sum of 2 would be 3/8.
C. The event of observing the sum of the face-up numbers to be at least 1 can happen in any case where at least one chip shows a face-up number of 1. There are multiple ways this could happen.
We can count all the favorable outcomes where the sum is at least 1:
(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 1, 1).
The number of favorable outcomes is 7. Since the sample space has 8 possible outcomes, the probability of observing a sum of at least 1 would be 7/8.
D. The probability of observing the sum of the face-up numbers equal to 2.5 is not applicable in this case because the sum of the face-up numbers must be a whole number since they can only take values of either 0 or 1. Thus, the probability of observing a sum of 2.5 would be 0.