The possible outcomes for rolling a number cube twice are:
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
This is a total of 36 different combinations.
There are 36 possible outcomes.
6 possible numbers to land on the first time, 6 possible numbers to land on the second time, 6x6=36
When a number cube is rolled twice, there are 36 possible outcomes. (1,1),(1,2),....(6,6). (3,3) occurs only once. Therefore, the probability of rolling a 3 both times is 1/36.
When rolling a number cube (a six-sided dice) twice, the sample space consists of all possible outcomes from both rolls. Since each roll has 6 possible outcomes, the total number of outcomes for rolling the number cube twice is 6 x 6 = 36. The sample space would be {1-1, 1-2, 1-3, ..., 6-5, 6-6} representing all possible combinations of the two rolls.
Well what does the spinner look like
There are 36 possible outcomes.
36
6 possible numbers to land on the first time, 6 possible numbers to land on the second time, 6x6=36
When a number cube is rolled twice, there are 36 possible outcomes. (1,1),(1,2),....(6,6). (3,3) occurs only once. Therefore, the probability of rolling a 3 both times is 1/36.
If a spinner has six possible outcomes, then there are 36 (62) permutations of outcomes from spinning it twice.
It is 0.722... recurring.
When rolling a number cube (a six-sided dice) twice, the sample space consists of all possible outcomes from both rolls. Since each roll has 6 possible outcomes, the total number of outcomes for rolling the number cube twice is 6 x 6 = 36. The sample space would be {1-1, 1-2, 1-3, ..., 6-5, 6-6} representing all possible combinations of the two rolls.
Well what does the spinner look like
The odds of corerectly predicting 1 number on a dice with 1 throw is 1 in 6. Predicting the same number to appear in two throws, its 1 in (6 * 6) 36 > Same deal with the coin, except less options, so 1 in (2 * 2) 4
There are four outcomes possible. Both even, both odd, and one of each twice. So, in one roll, it looks like your chances are 1 in 4. Is that what you are after?
The odds of of an odd number on the first role are 3/6 since the odds are 1,3, and 5 and there are 6 numbers. The odds of an even are the same. Since they are independent we can multiply the two probabilities and the answer is .5x.5=.25 If this is confusing, consider the sample space where I denote O for an odd number and E for an even number. In two rolls you can have: OO, OE,EO, OO So there are 4 possible outcomes. We are interested only in OE which is 1 of the 4 outcomes so the probability of this happening is 1 in 4 or .25.
Because 3/6 of the sides on a number cube have even numbers, the probability of rolling even on one number cube is 1/2(equivalent of 3/6). But since you're rolling twice, you multiply the probability of one by itself (therefore rolling 2 number cubes). So: 1/2x1/2=1/4 The probability of rolling an even number when a number cube is rolled twice is 1/4, 25%, or 1 out of 4.