Answer to Question #201045 in Statistics and Probability for Lou Robert Cuyos

Question

If one ball each is drawn from 3 boxes, the first containing 3 red, 2 yellow, and 1 blue, the second box contains 2 red, 2 yellow, and 2 blue, and the third box with 1 red, 4 yellow, and 3 blue. What is the probability that all 3 balls drawn are different colors? 

Answer 1

Let R denotes red ball

Y denotes yellow ball and

B denotes blue ball

So, according to question:

The possibilities are RYB , YRB , BRY , RBY , BYR , YBR

$P(RYB)=$ P[Red ball first box ; Yellow ball from second box ; Blue ball from third box]

$= \dfrac{3}{6}\times \dfrac{2}{6}\times \dfrac{3}{8}= \dfrac{9}{144}$

$P(YRB)=\dfrac{2}{6}\times \dfrac{2}{6}\times \dfrac{3}{8}=\dfrac{6}{144}\\\ \\P(BRY)=\dfrac{1}{6}\times \dfrac{2}{6}\times \dfrac{4}{8}=\dfrac{4}{144}\\\ \\P(RBY)=\dfrac{3}{6}\times\dfrac{2}{6}\times \dfrac{4}{8}=\dfrac{12}{144}\\\ \\P(BYR)=\dfrac{1}{6}\times \dfrac{2}{6}\times \dfrac{1}{8}=\dfrac{1}{144}\\\ \\P(YBR)=\dfrac{2}{6}\times \dfrac{2}{6}\times \dfrac{1}{8}=\dfrac{2}{144}$

$P(\text{three balls drawn are of different colors})=\dfrac{9+6+4+12+1+2}{144}=\dfrac{34}{144}=0.236$