suppose that 2 balanced dice are tossed repeatedly and the sum of the 2 uppermost faces is determined on each toss. what is the porbabilty that we obtain a sum of 3 before we obtain a sum of 7.
since obtain 3 and 7 are independent,
P( getting sum of 3) = 2/6 x 1/6
P ( getings sum of 7) = 3/6 x 1/6
so the answer should be 2/6 x 1/6 x 2/6 x 1/6
but my book says taht it is 1/4...
supposed it is to get a sum of 4 before a sum of 7, the method would be the same just that instead of 2/36 in the final equation, it will be 3/ 36 divide by 8/ 36 right?