Bunuel wrote:
Official Solution:
A jar contains only \(B\) blue balls, \(6B + 10\) yellow balls and \(2B+5\) green balls. If one ball is picked at random from the jar what is the probability of getting a blue or green ball?
A. \(\frac{1}{5}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{3}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
Total # of balls in the jar is \(B + (6B + 10) + (2B + 5) = 9B + 15\);
The probability of picking a blue or green ball is \(\frac{B+(2B + 5)}{9B + 15}=\frac{1}{3}\).
Answer: C
Hey Bunuel,
Is the trick to simplifying that expression into 1/3 just simply seeing that the ratio of top to bottom is 1:3? Sometimes I get caught up doing a problem and try to simplify algebraically without "zooming out" for a second to see that. Would you mind showing how you would simplify that algebraically just to satisfy my curiosity?