Carcass wrote:

Each block in a certain bucket is either red, yellow, or blue. There are half as many red blocks as blue blocks and 3 times as many yellow blocks as red blocks. If a block is withdrawn from the bucket at random, what is the probability that the block is blue?

A. \(\frac{1}{6}\)

B. \(\frac{1}{3}\)

C. \(\frac{1}{2}\)

D. \(\frac{2}{3}\)

E. \(\frac{2}{6}\)

Carcass..let me know my method is correct.

Given that there are half as many red blocks as blue blocks => 2R = B

and 3 times as many yellow blocks as red blocks => Y = 3R

Let's take some number that is divisible by both 2 and 3, here in this let's take B as 12, then R = 6

when R = 6 then Y = 18.

Total = 6+12+18 = 36.

P(B) = Blue / Total => 12/36 => 1/3

B.