Probablity problems when rolling two dice?

There are 36 combination of rolls (6*6). For simplicity ignore the first dice roll.

"what is the probability that the second dice has a higher score than the first dice?"

A roll of 1 is higher than nothing (0), a roll of 2 is higher than 1 (1), a roll of 3 is higher than 1,2 (2), etc. etc. Is the simplest way to explain it. Therefore 0+1+2+3+4+5 = 15/36(41.6%) and same logic vice versa for the second problem.

"what is the probability that both the dice end up having the same score?"

Six possible outcomes: double 1s, 2s, 3s, etc.

6/36 = 1/6 (16.6%)

The singular of dice is "die."

You need to figure that if die 1 has a 1, what is the probability that the second will have a 2, 3, 4, 5, 6? that would be 5/6.

If die 1 is a 2, what are the chances of getting a 3,4,5,or 6? That would be 4/6.

Continue and add up the probablilities.

Probablility that both dice would be the same? The probablility of a 1 on the first is 1/6 and a 1 on the second is 1/6, so the probability of 2 ones, is 1/6 x 1/6 or 1/36. This would be true for any of the six sides.

Remember, if you have two or more exclusive probabilties: if they are joined by "and," you multiply the chances of each; if they are joined by "or," you add them

First, decide which die is no 1 and which is no 2.

Work out all the possible throws.

1st =1, 2nd = 2,3,4,5,6: 5 possible throws

1st =2, 2nd = 3,4,5,6: 4 throws

1st =3, 2nd = 4,5,6: 3 throws

1st =4, 2nd = 5,6: 2 throws

1st =5, 2nd = 6 only: 1 throw

Total = 15 throws.

There are 36 throws in all (6 x 6)

So the prob that the 2nd die has a higher number than the first is 15/36 or 5/12.

The same answer applies if the first die is to have a higher number than the second. Again 5/12.

The probability that both show the same number is

6 in 36, as you can have 1-1, 2-2, 3-3, 4-4, 5-5, 6-6.

Thus 6 chances in 36, or 1 in 6.

1. Suppose the roots are a, b, and c. Then the factored form of the equation is (x - a)(x - b)(x - c) = 0. When you multiply the factiors out yoy get x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc. This shows that the negative product of the roots is the constant term. Therefore, the product of the roots is -12. (Check: the roots are 3, (-5 +/- sqrt(41))/2.) 2. The probability is 20/36 = 5/9. You can write out the 36 possibilities in the sample space and compute the difference of squares.