Hey,

Question asks the possibility of tossing the dice two times before tossing blue, in other words it asks. 1(all possible tosses) - the possibility of tossing the dice blue before the third toss. In this case there are 2 possibilities. You can toss the dice blue first and stop, the second possibility is that you toss the dice it is not blue at first and you toss it again and it is blue at the second.

A: Blue, Stop. Probability of this occurring is = 2 (number of blue faces) /8 (all faces)

B: Non-Blue, Blue, Stop. Probability of this occurring is= 6 (non-blue faces) /8 (all faces) x 2 (blues) /8 (all)

C: All possible toss combination = 1

Answer of the question is: C - A -B = 0.5625