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# How many times will the number 5 show up if a fair number cube rolled 300 times?

If the cube faces are numbered 1 through 6 then 5 will show up once every six times, so 300 rolls will show up a 5 about 50 times based on probability

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# How many times would you expect to roll a number less than 5 when you toss a die 360 times?

Since a cube has 6 sides, you have a 4 in 6 chance of getting a number less than 5 each time you roll the dice. You would take how many times you want to roll it (360), divide… it by the number of sides (6), and multiply it by 4 since you know you have a 4 in 6 chance. So it should be 240. This is strictly mathematical though and you should not consider it accurate since there are many variables when rolling it.

# A number cube is rolled three times. What is the probability of rolling a 2 each time?

the chances of rolling a 2 each time during 1 set of 3 rolls is -1 in 216- sets. i believe the formula is: 1 in 6 - the chance of rolling a"2" then multip…ly the chances for each separate roll 1/6 x 1/6 x 1/6 = 1/216

# What is the theoretical probability of rolling a number greater than a 1 on a fair number cube?

5/6 0.833333333 ect. 83%

# A number cube is rolled 2 times in a row What is the probability of rolling a 3 both times?

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 you roll a dice 300 times about how many times would you expect it to land on 5?

1/6 of all outcomes should be a 5. 300*1/6=50

# What is the probability of rolling a number divisible by 3 with two rolls of a fair number cube?

The probability of rolling a number divisible by three on a fair die (3 and 6) is 2 in 6, or 1 in 3, or about 0.3333. Since there are 36 permutations of rolling two dice, the …probability, then, of rolling (at least one) number divisible by three on two rolls of a die is 19 in 36, or about 0.5278. Simply count the number of desired results, and divide by the number of possible results, as I have done below. 11 12 13 x 14 15 16 x 21 22 23 x 24 25 26 x 31 x 32 x 33 x 34 x 35 x 36 x 41 42 43 x 44 45 46 x 51 52 53 x 54 55 56 x 61 x 62 x 63 x 64 x 65 x 66 x If, on the other hand, you are looking at the probability of rolling a sum divisible by three, consider ... 11 2 12 3 x 13 4 14 5 15 6 x 16 7 21 3 x 22 4 23 5 24 6 x 25 7 26 8 31 4 32 5 33 6 x 34 7 35 8 36 9 x 41 5 42 6 x 43 7 44 8 45 9 x 46 10 51 6 x 52 7 53 8 54 9 x 55 10 56 11 61 7 62 8 63 9 x 64 10 65 11 66 12 x ... which is 12 in 36, or 1 in 3, or about 0.3333.

# If you roll a fair number cube 30 times how many times would you expect to roll a odd number?

Type your answer here... 15

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# How many times is the number 300 mentioned in the bible?

19 times Gen_5:22; Gen_6:15; Jdg_7:6; Jdg_7:7; Jdg_7:8; Jdg_7:16; Jdg_7:22; Jdg_8:4; Jdg_11:26; Jdg_15:4; 1Ki_10:17; 1Ki_11:3; 1Ch_11:11; 1Ch_11:20; 2Ch_9:16; 2Ch_14:9; 2…Ch_35:8; Ezr_8:5; Est_9:15;

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# If You roll a fair number cube twice What is the probability of rolling two threes if the first roll is a 5?

2/36 what you rolled the first time has nothing to do with the next roll. It's an independent event.

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# When rolling a number cube once what is the probability of rolling a 5?

1 out of six

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# What is the probability of rolling a 5 on a number cube?

It is 1/6.

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# When rolling a number cube twice what is the probability of rolling the same number each time?

1 out of 6, if the highest number on the cube is 6. hope that helped!:) --- If only rolling ONE 6-sided die, and it can be any first roll that is matched by the second roll, t…he odds would be 1 in 6. It would not matter what the 1st roll came up with, so 6 in 6 rolls meet the 1st criteria. But the second roll could be any of the six sides, and only one of them would match whatever the number rolled first was. If however, rather than a cube you used a differently shaped die, with a different number on each of its faces, the odds would be one in however many sides the die has. When tossing 2 dice the problem can be more difficult to compute if satisfied by the total to be matched, or if the exact same #s need to be repeated. For example, a 1 and a 6 totals 7, which can also be satisfied by a 2 and 5, or 3 and 4, etc.