you roll two fair dice, a green one and a red one?
(a) are the outcomes on the dice independent?
(b)Find P(1 on green die and 2 on red die)
(c)Find P(2 on green die and 1 on red die)
(d)Find P(1 on green die and 2 on red die) or (2 on green die and 1 on red die)
a)yes
b)p(1)*(p2)=1/6*1/6=1/36
c)p(2)*(p1)=1/36
d)1/36=1/18
Agree with all answers
(a) Are the outcomes on the dice independent?
Well, when it comes to dice, they don't really care about each other, so yes, the outcomes on the dice are usually independent. I mean, they don't have any pre-determined agreement like "You roll a 5, and I'll roll a 2." So, they're just doing their own thing.
(b) Find P(1 on green die and 2 on red die)
Ah, the classic case of dice harmony. To figure out the probability of this happening, you'll need to multiply the individual probabilities together. So, P(1) on the green die is 1/6, and P(2) on the red die is also 1/6. Multiply them, and you get 1/36. Voila, you have your probability!
(c) Find P(2 on green die and 1 on red die)
Oh, this time the dice want to switch it up, going for a different roll. The probability of getting a 2 on the green die is still 1/6, while the probability of getting a 1 on the red die is also 1/6. Multiply them together, and you get another 1/36. Looks like they're trying to keep the universe balanced or something.
(d) Find P(1 on green die and 2 on red die) or (2 on green die and 1 on red die)
Well, well, well, what do we have here? Looks like a little case of "either/or." To find the probability of either event happening, you'll just add up their individual probabilities. We already know that the probability of getting a 1 on green and a 2 on red is 1/36. And guess what? The probability of getting a 2 on green and a 1 on red is also 1/36! Add them up, and you get a grand total of 1/18. Tada! Clown math strikes again.
Remember, the key to understanding probabilities is to have a little fun with it. And if all else fails, just blame it on the clowns. We're here for entertainment, after all!
To answer these questions, we need to understand a few concepts. When we roll two fair dice, the outcomes on each die are independent. This means that the outcome of the green die does not affect the outcome of the red die, and vice versa.
Now let's go through each question step by step:
(a) Are the outcomes on the dice independent?
To determine if the outcomes are independent, we need to check if the outcome of one die influences the outcome of the other. In this case, the rolling of the green die does not affect the rolling of the red die, and vice versa. Therefore, the outcomes on the dice are indeed independent.
(b) Find P(1 on green die and 2 on red die)
To find the probability of getting a 1 on the green die and a 2 on the red die, we need to multiply the individual probabilities of each event. The probability of getting a 1 on a fair die is 1/6, and the probability of getting a 2 is also 1/6. Therefore, the probability of getting a 1 on the green die and a 2 on the red die is (1/6) * (1/6) = 1/36.
(c) Find P(2 on green die and 1 on red die)
Using the same logic as the previous question, the probability of getting a 2 on the green die is 1/6, and the probability of getting a 1 on the red die is also 1/6. Therefore, the probability of getting a 2 on the green die and a 1 on the red die is (1/6) * (1/6) = 1/36.
(d) Find P(1 on green die and 2 on red die) or (2 on green die and 1 on red die)
To find the probability of either event occurring, we need to add the probabilities of each event separately. Using the probabilities calculated earlier, the probability of getting a 1 on the green die and a 2 on the red die is 1/36, and the probability of getting a 2 on the green die and a 1 on the red die is also 1/36. Therefore, the total probability is (1/36) + (1/36) = 2/36 = 1/18.
Remember that when calculating probabilities, we always multiply the probabilities for independent events and add the probabilities for mutually exclusive events. In this case, the outcomes on the dice are independent, which allows us to calculate the probabilities using simple multiplication.