Question Video: Determining the Probability of an Event Involving Dice | Nagwa Question Video: Determining the Probability of an Event Involving Dice | Nagwa

# Question Video: Determining the Probability of an Event Involving Dice Mathematics

Suppose you roll two number cubes, where each cube has the numbers 1, 2, 3, 4, 5, and 6 on its faces. Determine the probability that the sum of the two numbers rolled is less than 12.

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### Video Transcript

Suppose you roll two number cubes, where each cube has the numbers one, two, three, four, five, and six on its faces. Determine the probability that the sum of the two numbers rolled is less than 12.

So, let’s have a look at this question and the two cubes. We’re told that they have the numbers one, two, three, four, five, and six on each cube. And when we roll our cubes, we’re interested in the number at the top. In the picture, here, the numbers would be one and six. Let’s have a look at the different options we could get when we roll these two cubes. We could have a one on each cube, a one and a one. We could also have one on the first one and a two on the second cube. We could also have a two on the first cube and a one on the second cube.

We could carry on listing all the different outcomes, but it might be more sensible to put it into a neater format. So, let’s put the information into a table with one of the cubes along the rows and one of the cubes on the columns. Since we know that each cube has the numbers one through six, then we can write these values in the row and the column. The first empty cell in our table would signify the event when we have one on cube one and one on cube two. The second cell would be a one on cube one and a two on cube two. And we could continue to fill in the remainder of the row with one, three; one, four; one, five; and one, six.

For the next row of the table, we would fill in for whenever we have a two on cube one and then a one through six on cube two. And we could continue filling in the values across the table. So, now, we have listed all the different possible outcomes from rolling our two cubes. We can see in the question that we’re interested in the sum of the two numbers. So, if we look at the first cell in our table, we have the event of one and one on the cubes. The sum of these two values would be two.

In the next cell, we would have a one and a two on our cubes, and the sum of those two numbers would be three. In the same way, a one and a three on the cubes would sum to give us a value of four. We can then continue adding our values along the first row. In the next row, we’re adding two and one to give us three. And then, we can continue filling in the table to give us a table where we have added the values on cube one and cube two.

So, now, let’s see how many times we would get a sum that’s less than 12. We can see in the first row that there are six different possibilities that will give us a sum of less than 12. In the second row, there are also six values with a sum less than 12. And the next three rows after that will also give us six different values. In the last row, however, there are only five values less than 12, seeing as the 12 wouldn’t count as one of them.

So, then, to find the total number of times we would get a sum of less than 12, we add our five number sixes and our five, which gives us 35. We could also have got this value by noting that there are 36 total possible outcomes. Then, since there’s only one occurrence where it’s 12 or more, this means we would work out 36 minus one, giving us 35. And now, we can say that the number of times that the sum is less than 12 is 35 times.

So, now, to find the probability of the sum of the two numbers being less than 12, we can use the formula that the probability of an event — in this case, the experimental probability — is equal to the number of times the event occurs over the total number of outcomes. So, to find the probability of a sum less than 12, we work out the number of times that the sum is less than 12 and divide it by the total number of outcomes. So, since we know that there are 35 times when the sum is less than 12, we can substitute in that value. And we know that there are 36 different possible outcomes. So, we can write our answer as a fraction as 35 over 36. And since we can’t simplify this fraction any further, this will be our final answer.

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