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?
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 ...
A. To write the sample space, we need to list all the possible outcomes.
Each chip can have either a 0 or a 1 face-up. Since all three chips are identical, we can write the sample space as a combination of the outcomes of each chip. Let's represent the outcomes as (chip1, chip2, chip3).
The possible outcomes for each chip are:
- Chip 1: (0, 1)
- Chip 2: (0, 1)
- Chip 3: (0, 1)
Therefore, the sample space for tossing three chips simultaneously is:
{(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)}
B. To calculate the probability of observing the sum of the face-up numbers equal to 2, we need to determine the number of favorable outcomes and divide it by the total number of possible outcomes.
The favorable outcomes are: {(1, 1, 0), (1, 0, 1), (0, 1, 1)}
Therefore, there are 3 favorable outcomes.
Since the sample space has 8 total outcomes, the probability is calculated as:
P(sum = 2) = Number of favorable outcomes / Total number of outcomes = 3/8
C. To calculate the probability of observing the sum of the face-up numbers to be at least 1, we need to determine the number of favorable outcomes where the sum is 1 or greater and divide it by the total number of possible outcomes.
The favorable outcomes are: {(0, 0, 1), (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)}
Therefore, there are 7 favorable outcomes.
Since the sample space has 8 total outcomes, the probability is calculated as:
P(sum ≥ 1) = Number of favorable outcomes / Total number of outcomes = 7/8
D. The sum of face-up numbers cannot be 2.5 since the numbers on each chip is either 0 or 1. Therefore, the probability of observing the sum of the face-up numbers equal to 2.5 is 0.