We make successive throws of two balanced dice of six faces and we are interested in finding the probability of the event that the sum of 5 (the faces of the two dice) will appear before the sum 7. We assume that the throws are independent.
First calculate the probability of the event E(n): In the first n - 1 throws, neither the sum 5 nor the amount 7 occurred, and in the n throw the amount was 5.
The same question, but replacing 5 with 2.
I'm stuck . I have found that the sum 5 can appear in 4 cases : (1,4);(4,1);(2,3);(3,2).
And 7 in 6 cases : (1,6);(6,1);(2,5);(5,2);(3,4);(4,3).
All possible cases are 6*6=36 .
I found the probability of sum 7 : 6/36 = 1/6 = 0,16 And the probability of sum 5 : 4/36 = 1/9 = 0,11
Now I'm stuck and I don't even know if all I did here it's correct. If I understand how to calculate solution for 1, I can do alone for 2. Please help me !