Two faces of a six-sided die are painted red, two are painted blue, and two are painted yellow. The die is rolled three times, and the colors that appear face up on the first, second, and third rolls are recorded. Find the number of elements in the sample space whose outcomes are all possible sequences of three rolls of the die. (a) Find the probability of the event that exactly one of the colors that appears face up is red. (b) Find the probability of the event that at least one of the colors that appears face up is red.

Sample space = 27 possible outcomes

a) P(just one red) = 4/9

b) P(at least one red) = 19/27

Step-by-step explanation:

Each die has 3 possible outcomes (red, blue or yellow), so the sample space of the number of possible outcomes for rolling three dice is 3^3 = 27

a)

The probability of the face up color being red is 2/6 = 1/3, and the probability of it not being red is 4/6 = 2/3.

The die with the face up red can be any of the three dice, so we also need to multiply the probabilities by a combination of 3 choose 1. So the final probability is:

P = C(3,1) * (1/3) * (2/3)^2 = 3 * (1/3) * (4/9) = 4/9

b)

To find the probability of at least one face up being red (event A), we can find the probability of the complementary case, that is, no one of the three die being face up red (event A'), and then we calculate P(A) + P(A') = 1.

In the case where no face up are red, we have the probability of 2/3 for each die, so we have:

P(A') = (2/3)^3 = 8/27

Then we can find P(A):

P(A) = 1 - 8/27 = 19/27

8

step-by-step explanation:

this is a subtraction problem.

10 - 2 = 8

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )