Respuesta :
Answer: [tex]\text{P(spiral-patterned paper and a green)}=\frac{1}{12}[/tex]
Step-by-step explanation:
Given: The number of patterns in wrapping paper = 4
The number of colors available for ribbon = 3
Then, the total number of types of ribbons are in store =[tex]4\times3=12[/tex]
Also, the number of type of ribbon has spiral-patterned paper and a green in color =1
So, if Gary randomly selects a paper pattern and a ribbon color, then the probability that he chooses spiral-patterned paper and a green ribbon is
[tex]\text{P(spiral-patterned paper and a green)}=\frac{1}{12}[/tex]
Answer:
Given: The number of patterns in wrapping paper = 4
The number of colors available for ribbon = 3
Then, the total number of types of ribbons are in store =
Also, the number of type of ribbon has spiral-patterned paper and a green in color =1
So, if Gary randomly selects a paper pattern and a ribbon color, then the probability that he chooses spiral-patterned paper and a green ribbon is