To solve the question above, we can follow a few methods, depending on your knowledge of mathematics.

A. We can use probability formulas to understand the state of the bag in each turn.

B. We can write down and draw the state of the bag in each turn.

When it comes to this exam, I highly suggest using a combination of both. Understanding the mathematical idea of the question while also drawing and writing down the steps will allow you to create a clear picture of the question and compare it to the options provided.

Step 1) Three red balls, three yellow balls, and one green ball are placed in a bag - There is \frac{3}{7} probability of pulling out a red ball, \frac{3}{7} of pulling out a yellow ball, and \frac{1}{7} of pulling out a green one, **during the first round.**

*You pull out one red ball and then one green ball*

Step 2) You have two red balls \frac{2}{5} , three yellow balls \frac{3}{5} , and no green ball \frac{0}{5} in your bag.

A - The next ball could be any one of red, yellow, or **green**. We don’t have any green balls left.

B - The next ball will definitely be yellow - That cannot be true, as we have \frac{2}{5} chance of pulling out a red ball, as mentioned in the quick calculation we made.

C - The next two balls cannot both be red. - This is a possibility, as we already calculated that we have two red balls left.

D - At least one of the next three balls must be yellow. - It **must** be true as we only have two red balls left in the bag and **all of the rest** yellow.

E - At least one of the next three balls must be red. - As we have three yellow balls in the bag, it is possible that we will pull all of them out in the next three rounds.

**Tip:** This question might look straightforward when solving it at home, but you have to consider the fact that it will be a bit different on the exam. Some simple questions with a lot of info and steps might take a bit longer to solve, so I highly suggest always approaching these questions, knowing which actions you will follow while solving them.