probability of rolling dice?

Question

a pair of standard dice, one red and one green, is rolled. what is the probability that
a) the sum of the pips on the top faces is 6 or 8?
b) the sum of the pips on the top faces is 4 or a "double" is rolled?
c) the red die shows 4 pips on the top face or the green die shows a 5 on the top face?

Answer 1

There are 6 x 6 = 36 possible rolls.

(a) To make six, you can roll 1,5....2,4...3,3...4,2...5,1. To make 8 you can roll 2,6...3,5...4,4...5,3...6,2. So there are 10 ways in total to make 6 or 8, for a probability of 10/36.

(b) There are obviously 6 doubles. To make a sum of 4 there are 3 ways: 1,3...2,2...3,1. But the 2,2 is one of the doubles, so the total is 8 ways to get either a double or sum of 4. Probability is this 8/36.

(c) There is a 5/6 probability that the red does NOT roll a 4. There is also a 5/6 probability that the green does NOT show a 5. The probability of both of these events occurring -- i.e. not a red 4 and not a green 5 -- is the product, namely 25/36. Thus the probability that one or the other (or both) *does* occur is 1 - 25/36 = 11/36.

Answer 2

Since a standard die has 6 sides, each number has a 1/6 chance of coming up. When there are multiple sides that will fulfill your query (ie Come up 4 or 6), then you add the chances of each event (4 = 1/6 + 6 = 1/6..... 1/6 + 1/6 = 2/6 or 1/3). When there are two dice, and each have specific events that will fulfill the query (4 and 6 at same time) then you multiply the chance of each event (1/6 * 1/6 = 1/36).
The simplest way to figure out your questions is to decide how many (and which) events will fulfill the query, probably by making a quick table, then doing the math operations as above. These look like hw problems, so I won't do them for you, but I hope this helps. Good Luck!

Answer 3

each roll of the die is a Bernoulli trial to that end. via definition a Bernoulli trial is an experience that has 2 accessible outcomes, a fulfillment or a failure. here a fulfillment is rolling a three with fulfillment danger of a million/6; a failure is rolling something different than a three. each and each roll is an self sufficient Bernoulli trial. understanding that the 1st ten rolls are would not help us to comprehend what the eleventh or twelfth roll would be. Now, i will amend this quite. If we anticipate the die is truthful then the danger is a million/6 that the 1001st roll is a three. If the 1st 1000 rolls tutor an experimental danger of a three to be, say a million/4, i.e., the die isn't truthful, then we would evaluate the danger of the 1001st roll being a three to be a million/4.

Answer 4

a) you have 36 combinations of 2 six-sided dice.
5 of the combinations total 6 : [(3,3),(1,5),(5,1),(2,4),(4,2)]
5 of the combinations total 8 : [(4,4),(2,6),(6,2),(5,3),(3,5)]
so
a: 10/36 = 5/18

b) sum = 4 with no doubles = (1,3)(3,1) so 2 combos. there are 6 combos of doubles so 8 total combos of 4 or doubles. 8/36 = 2/9

Answer 5

For 6, 1,5 or 2,4 or 3,3 or 4,2 or 5,1
for 8, 2,6 or 6,2
The answer is 7/36