When rolling a dice two times, what is the probability of having a number greater that 3?

There is a 50% chance of getting this result each time that you roll the die.  Here is why:

First, I am assuming that you are using only one die.  Second, I am assuming that the die has six sides and that the numbers on those sides are 1 through 6.

This means that each time that you roll the die, you get a 50% chance of getting a number greater than 3.  This is because 3 of the 6 sides die have numbers greater than 3.

Each time that you roll the die, you have the same probability of having a number higher than 3.  So each time you roll it, you have a 50% chance.

The question is ambiguous. If we need to calculate the probability of getting a number more than 3 in both the rolls of dice, then the answer #1 posted above by hala 718 is correct. However, the question could also mean that we get a number greater than 3 in at least one of the two rolls. In this case the probability is calculated as the probability of not getting a number less than 3 in both the rolls. We calculate this as follows.

Probability of getting a number equal to or less than three in any one roll of dice:

= probability of getting 1, 2, or 3 out of the total six possible numbers.

= 3/6 = 1/2

Then:

Probability of getting a number equal to or less than three in both of two roll of dice:

= (1/2)x(1/2) = 1/4

Probability of getting a number more than three in either one or bot of the roll of dice:

= Probability of not getting a number equal to or less than three in both of two roll of dice.

= 1- (Probability of getting a number equal to or less than three in both of two roll of dice)

= 1 - 1/4 = 3/7 = 0.75

Answer:

Probability of getting a number greater than three at least once in two roll of dice = 0.75

Let x or y be the number coming up when you throw a dice. x = {1,2,3,4,5,6) and y = (1,2,3,4,5,6}. Probabilty of obtaining a number greater than 3, or  P(x+y >3) =1 -  P(x+y < =3 ).

Now x+y  could be (1,1), (1,2),.....(6,6) in 36 ways.

x+y <=3 could be in one of the (1,1), (1,2) , (2,1) 3 ways. Therefore, P(x+y <= 3) = 3/36. =1/12.

Therefore, P(x+y >= 3) = 1 - P(x+y <=3) = 1 -1/12 = 11/12.