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2 years ago

Please try your best and show your work and I will give brainliest!

Mathematics

1 answer:

dimulka [17.4K]2 years ago

5 0

Answer:

Part A: impossible

Part B: Either equal or blue

Part C: 9 green and 2 blue were added

Step-by-step explanation:

Part A:

The only colors included in this problem are red, blue, and green. There is no black colored pencil, therefore, it is impossible to get one from the box.

Part B:

I'm not sure what you're asking in this question, but I will give you the two choices. If it is before the additional 11 colored pencils are added to the box, the chance of drawing a red and the chance of drawing a blue will be equal, because both of them have 11 of each color. If it is after the additional 11 colored pencils are added to the box, then the chance of drawing a blue colored pencil will be greater than the chance of drawing a red colored pencil. After the 11 colored pencils are added, there are 13 blue and 11 red. The blue is greater.

Part C:

The least number of green colored pencils added has to be 9, because the chance of drawing a green pencil is now greater than the chance of drawing a red pencil. If we add 8 more green pencils, the likelihood would be the same. Therefore, the number of green colored pencils added has to be at least 9. If we have the last 2 colored pencils be blue, then there would be 11 red, 13 blue, and 12 green. This fits all the conditions, therefore, adding 9 green colored pencils and 2 blue colored pencils is the answer.

I hope this helps and please mark me as brainliest!

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3 years ago

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