# One dice rolled twice

What is the probability of one dice rolled twice that will result in exactly one three being rolled?

Also what is the probability of at least one three being rolled (one dice rolled twice)? I have the answers, I just need to understand it…

The answer for exactly one three comes up is 5/18 and at least one three comes up is 11/36. How did they get 5/18?

*Question posted by: atukomeyer*

## Answer

Let’s break down the probabilities for the two scenarios you’ve mentioned:

### Exactly one 3 being rolled (one dice rolled twice):

There are two possible ways for this to happen: either you roll a 3 on the first roll and not on the second, or you don’t roll a 3 on the first roll but do on the second. We’ll calculate the probabilities for each case and then add them together.

**Case 1**: Roll a 3 on the first roll and not on the second. The probability of rolling a 3 on a single roll is 1/6 (since there’s one 3 on a 6-sided die). The probability of not rolling a 3 on the second roll is 5/6 (as there are five other outcomes). So, the probability for this case is (1/6) * (5/6) = 5/36.

**Case 2**: Don’t roll a 3 on the first roll, but roll a 3 on the second. The probability of not rolling a 3 on the first roll is 5/6. The probability of rolling a 3 on the second roll is 1/6. So, the probability for this case is (5/6) * (1/6) = 5/36.

Now, we add the probabilities for both cases: 5/36 + 5/36 = 10/36, which simplifies to 5/18. That’s the probability of rolling exactly one 3 in two rolls of a single die.

### At least one 3 being rolled (one dice rolled twice):

An easier way to calculate this probability is to find the probability of NOT rolling a 3 in either roll and then subtracting that from 1.

The probability of not rolling a 3 in a single roll is 5/6. The probability of not rolling a 3 in both rolls is (5/6) * (5/6) = 25/36.

Now, subtract this probability from 1 to find the probability of at least one 3 being rolled: 1 – 25/36 = 11/36.