Since the problem reference unknown values and asks to know the distribution of said values, plugging in an easy number is a very viable tactic. In this case, we first want to locate the foundation variable dictating all aspects of the problem. Here, we see that the blue blocks dictate the red, which in turn dictate the yellow. So, plug in an easy value for the blue blocks such as blue = 4.

If there are 4 blue, and half as many red as blue, then there are 2 red blocks. Finally, if there are three times as many yellow as red blocks, then there are 6 red blocks.

Now, to determine the probability of selecting a blue block, remember that probability is defined as desired possibilities / total possibilities. In this case, there are 4 desired blue blocks out of a total of 4 + 2 + 6 = 12 blocks. Therefore, the probability of selecting a blue is 4/12 or 1/3, so select choice B.

in both equation red block is common, can't we make single ratio for B:R:Y.
if we do this B:2R and Y:3R,right?
this would led to 3B:6R:2Y?

yes but you have to be careful to make the ratios on the same level

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