Roll a fair 6 sided die twice, What is the probability that one or both rolls are 6?
This question is confusing me on a conceptual level, specifically the "or both rolls."
With out that, the question is what is the probability of getting a 6 by rolling a die twice, which is of course 11/36.
The problem I am having is I don't understand how the "or both" factors in. If you roll a 6 on the first and second roll (1/36 chance) is that counted separately than getting a 6 on the first or second roll?
And if so, would the answer be 1/6 + 1/6 + 1/36 = 13/36?
And that leads into my second question:
What is the probability that one or both rolls are even numbers (2, 4 or 6’s)?
IF the above work is correct, then by the same logic the chance of the first roll being even is 1/2 + the chance of the second roll being even which is 1/2 + the chance of them both being even which is 1/4. This adds up to >1 which is obviously impossible, but I don't know what is going wrong.
Thanks.
Best Answer
Almost. As you argued, probability to hit a 6 on each roll is 1/6, but the sum overcounts rolling 6 twice, so the actual probability of rolling 6 on either one or both roll is $$ \frac{1}{6} + \frac{1}{6} - \frac{1}{36} = \frac{11}{36}. $$ Can you take it from here?