The table below shows the number of pencils and of each color for all 25 pencils in the box it likelihood of an event a crime may be described as certain more likely equally likely less likely or impossible. part A: the pencils have all been placed in the box compare the likelihood of drawing a red pencil to the likelihood of drawing a blue pencil from the box show your work and or explain your reasoning.

part b: the original 25 colored pencils have all been placed back in the box in another 11 colored pencils have been out of the box the new pencils are either red blue or green the likelihood of drawing a green pencil is greater than the likelihood of drawing red pencil but less than the likelihood of drawing a blue pencil how many colors of each pencil were added to the box explain how you know that your answer is the only possible outcome that meets all the conditions. ILL GIVE BRAINLIEST AND NO LINKS​

Step-by-step explanation:

Part A : Impossible

Part B : Equally likely

Part C: 9 Green balls, no red ball and 2 blue balls are added to the box.

Hope the explanation is helpful to you.

Step-by-step explanation

25 Pencils in the box are as follows:

Red Color - 11 pencils, Blue Color -11 pencils, Green Color - 3 pencils.

Part A : A single pencil is drawn from the box.

The likelihood that the pencil is Black is Impossible.

This is because there is no Black pencil in the box. The pencils are of Red, Blue or green color. So the event of drawing a Black pencil from the box is not possible.

Part B: The pencils have all been placed in the box and a single pencil is drawn from the box.

The likelihood of drawing a red pencil  from the box is 11/25.

The likelihood of drawing a blue pencil from the box is 11/25.

The two cases are thus equally likely.

Part C: The original 25 colored pencils have all been placed back in the box and another 11 colored pencils have been added to the box. The new pencils are either red, blue or green. The likelihood of drawing a green pencil is now greater than the likelihood of drawing a red pencil, but less than the likelihood of drawing a blue pencil.

When there were 25 pencils in the box, the likelihood of drawing a green pencil was (3/25) = 0.12 and that of drawing a red pencil was (11/25) = 0.44.

The likelihood of drawing a green pencil was less than the likelihood of drawing a red pencil.

When 11 pencils are added, total pencils in the box = 25 +11 = 36.

Let no red pencil be added. Then likelihood of drawing a red pencil  = (11 / 36) = 0.31.

Let x be the number of green pencils added. Then total number of green pencils in the box = x + 3.

For  likelihood of drawing a green pencil to be greater than the likelihood of drawing a red pencil,  (x+3)/(36) > (11/36).

Therefore (x+3) > 11

      Thus x > 8.

Therefore at least 9 pencils  out of the 11 pencils added  to the box should be green, for  likelihood of drawing a green pencil to be greater than the likelihood of drawing a red pencil.

There are 6 possible cases:

Case (i) 9 pencils are Green, 0 are Red and 2 are Blue.

Then likelihood of drawing a green pencil = (9+3)/36 = 0.33

         likelihood of drawing a Red pencil = (11)/36 = 0.31

         likelihood of drawing a Blue pencil = (2+11)/36 = 0.36

         The likelihood of drawing a green pencil is now greater than the likelihood of drawing a red pencil, but less than the likelihood of drawing a blue pencil.

Case (ii) 9 pencils are Green, 1 is Red and 1 is Blue.

Then likelihood of drawing a green pencil = (9+3)/36 = 0.33

         likelihood of drawing a Red pencil = (1+11)/36 = 0.33

         likelihood of drawing a Blue pencil = (1+11)/36 = 0.33

         All three are equally likely. so this case is not possible.

Case (iii) 9 pencils are Green, 2 are Red and 0 are Blue.

Then likelihood of drawing a green pencil = (9+3)/36 = 0.33

         likelihood of drawing a Red pencil = (2+11)/36 = 0.36

         likelihood of drawing a Blue pencil = (11)/36 = 0.31

         The likelihood of drawing a green pencil is less than the likelihood of drawing a red pencil. This case is not possible.

Case (iv) 10 pencils are Green, 0 are Red and 1 is Blue.

Then likelihood of drawing a green pencil = (10+3)/36 = 0.36

         likelihood of drawing a Red pencil = (11)/36 = 0.31

         likelihood of drawing a Blue pencil = (1+11)/36 = 0.33

         The likelihood of drawing a green pencil is greater than the likelihood of drawing a blue pencil. This case is not possible.

Case (v) 10 pencils are Green, 1 is Red and 0 are Blue.

Then likelihood of drawing a green pencil = (10+3)/36 = 0.36

         likelihood of drawing a Red pencil = (1+11)/36 = 0.33

         likelihood of drawing a Blue pencil = (11)/36 = 0.31

         The likelihood of drawing a green pencil is greater than the likelihood of drawing a blue pencil. This case is not possible.

Case (vi) 11 pencils are Green, 0 are Red and 0 are Blue.

Then likelihood of drawing a green pencil = (11+3)/36 = 0.39

         likelihood of drawing a Red pencil = (11)/36 = 0.31

         likelihood of drawing a Blue pencil = (11)/36 = 0.31

         The likelihood of drawing a green pencil is greater than the likelihood of drawing a blue pencil. This case is not possible.

Thus the only possible case is Case (i) 9 pencils are Green, 0 are Red and 2 are Blue.

3 is greater then both of those numbers

there is no picture in order for this question to be answered