Double dice question with respect of prime number, even number and odd number
Answers
People have already said the mathematical way to solve this. I’ll say the slower, less technical ( and less efficient time-wise ) way, mostly because I believe that if you can’t grasp the reasoning behind the calculation, you’ll have problems with this kind of thing even with equations.
You throw two dice, presumably fair and 6 sided. Each side has a 1/6 chance to get any given number.
What you want is a combination of two specific numbers in each die. This means, probability wise, that you can have any number in the first die, as long as the second one is also the same. So, the probability of getting both die land at the same number ( a double ) would be 1 * 1/6, or 1/6.
However, you don’t want any number, you want a even number. There are 3 even numbers in a die, so instead of having any number in the first die, you want 3/6 numbers in that first die, but the probability in the second hasnt changed, after all you still want the second die to match the first.
So, the probability is now 3/6 * 1/6. Or 3/36. Or 1/12.