A fair dice is rolled 3 times, The probability of the product of the three outcomes is a prime number is?
The products which give a prime number I found out to be only 4.
However for the total number of products, I'm confused.
I know that I just can't use 216 as total as combinations such as 6×3×1=3×3×2=6×1×3 are counted not only twice but thrice!!
Also there are many products with such multiple countings.
So I what I am asking for is to a way to count the total possible products without any repitition.
Thanks.
It is best to use the natural sample space of $6^3$ elements. That is because they are equally likely, which turns our problem into a straightforward counting problem.
If we use this sample space, there are precisely $9$ "favourables".