Question

Tree diagram:
Emily has a box with 4 different colored tiles: one red, one green, one blue and one yellow. If he draws one of the pieces without looking, what is the probability of drawing the green before the red?

Answer 1

Answer: $\dfrac{1}{12}$

Step-by-step explanation:

Given: Emily has a box with 4 different colored tiles: one red, one green, one blue and one yellow.

We assume that repetition is not allowed

Total number of ways to draw two tiles = $^4P_2=\dfrac{4!}{(4-2)!}$  [By permuattaions]

$=\dfrac{4\times3\times2}{2}=12$

Favourable outcome = First green then red  (only one way)

So, the probability of drawing the green before the red $=\dfrac{\text{favorable outcomes}}{\text{Total outcomes}}$

$=\dfrac{1}{12}$

hence, the required probability =$\dfrac{1}{12}$