Lydia tosses two six-sided number cubes.
a) list the sample space
b) what is the probability of Lydia rolling pairs of the same number?
c) what is the probability that at least one number will be a 3
a) The sample space of tossing two six-sided number cubes is the set of all possible outcomes, which consists of all pairs of numbers that can be obtained by rolling two six-sided dice. The sample space can be written as:
{ (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) }
b) To find the probability of rolling pairs of the same number, we need to count the number of outcomes in the sample space where the two dice show the same number. There are 6 such outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Therefore, the probability of rolling pairs of the same number is:
P(pairs) = number of outcomes with pairs / total number of outcomes
= 6 / 36
= 1/6
c) To find the probability that at least one number will be a 3, we can count the number of outcomes where at least one of the dice shows a 3 and divide by the total number of outcomes. There are 11 outcomes where at least one die shows a 3: (1,3), (2,3), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,3), (5,3), and (6,3). Therefore, the probability that at least one number will be a 3 is:
P(at least one 3) = number of outcomes with at least one 3 / total number of outcomes
= 11 / 36
a) To find the sample space, we need to list all possible outcomes of tossing two six-sided number cubes. Each cube can show numbers from 1 to 6. So, the sample space for tossing two number cubes can be expressed as a set of ordered pairs, where each pair represents the outcome of the two cubes.
Sample space = {(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)}
b) To find the probability of rolling pairs of the same number, we need to count the number of favorable outcomes (where both numbers are the same) and divide it by the total number of possible outcomes in the sample space.
Favorable outcomes: {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
Total outcomes: 36
Probability = Number of favorable outcomes / Total outcomes
Probability = 6 / 36
Probability = 1/6
So, the probability of Lydia rolling pairs of the same number is 1/6.
c) To find the probability that at least one number will be a 3, we need to count the number of favorable outcomes. In this case, we need to consider all outcomes where at least one of the numbers in the pair is 3.
Favorable outcomes: {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (1,3), (2,3), (4,3), (5,3), (6,3), (3,3)}
Total outcomes: 36
Probability = Number of favorable outcomes / Total outcomes
Probability = 12 / 36
Probability = 1/3
So, the probability that at least one number will be a 3 is 1/3.
a) The sample space is the list of all possible outcomes when tossing two six-sided number cubes.
Sample Space: {(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)}
b) To calculate the probability of Lydia rolling pairs of the same number, we need to determine the number of favorable outcomes (the pair of the same number) and divide it by the total number of possible outcomes.
Favorable outcomes: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Total possible outcomes: 36 (as there are 6 possible outcomes for each cube)
Probability = Number of favorable outcomes / Total possible outcomes
Probability = 6 / 36
Probability = 1/6
Therefore, the probability of Lydia rolling pairs of the same number is 1/6.
c) To calculate the probability that at least one number will be a 3, we need to determine the number of favorable outcomes (where at least one number is 3) and divide it by the total number of possible outcomes.
Favorable outcomes: {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3), (6, 3)}
Total possible outcomes: 36
Probability = Number of favorable outcomes / Total possible outcomes
Probability = 11 / 36
Therefore, the probability that at least one number will be a 3 is 11/36.