### Video Transcript

A bag contains a total of 12 marbles. Four are green and the rest are yellow. Madison randomly takes a marble from the bag, records its color, and does not replace it. She then repeats this process for a second time. She draws the following tree diagram. There are three parts to this question. Write the values of the probabilities ๐ด, ๐ต, ๐ถ, and ๐ท in the probability tree. Give you answers as unsimplified fractions. Calculate the probability of choosing two marbles that are of the same color. Give your answer as a simplified fraction. Calculate the probability of choosing one marble of each color. Give your answer as a simplified fraction.

We will begin this question by looking at the first part which asks us to complete the tree diagram. We are told in the question that there are 12 marbles in the bag and four of them are green. The rest of the marbles are yellow, which means there must be eight of them, as 12 minus four is equal to eight. This is where the fraction eight twelfths or eight out of 12 has come from, as the probability of selecting a yellow marble is eight out of 12. As there are four green marbles, the probability that the first marble selected is green is four out of 12. It is also worth recalling at this stage that any branches coming out of the same point on a tree diagram must sum to one. Four out of 12 plus eight out of 12 is equal to 12 out of 12 or one whole one.

Letโs now consider the top half of the second part of the diagram. A green marble has already been chosen, which means that there are 11 marbles left in the bag of which three are green. This means that the probability that the second marble is also green is three out of 11 or three elevenths. After selecting a green marble with the first selection, there are still eight yellow marbles left in the bag. This means that the probability that the second marble is yellow if the first marble is green is eight out of 11 or eight elevenths. Once again, these two probabilities must sum to one. Three elevenths plus eight elevenths is equal to one.

We can now focus on the bottom half of the second part of the tree diagram. We have already been told that the probability that the second marble is green if the first marble is yellow is four out of 11 or four elevenths. As four elevenths plus seven elevenths is equal to one, the probability that the second marble is yellow is seven elevenths. This makes sense, as if the first marble selected was yellow, there would be seven yellow marbles left out of 11 in total. The four fractions in their unsimplified forms are ๐ด is equal to four twelfths, ๐ต is equal to three elevenths, ๐ถ is equal to eight elevenths, and ๐ท is equal to seven elevenths.

The second part of our question wants us to calculate the probability of choosing two marbles that are the same color. This can happen in two ways, firstly, by selecting a green marble, then another green marble or by selecting a yellow marble followed by another yellow marble. We recall that when dealing with probability tree diagrams, the AND rule means multiply. This means that to calculate the probability of the first marble being green and the second marble being green, we need to multiply four twelfths by three elevenths. Multiplying the numerators gives us 12, and multiplying the denominators gives us 132. The probability of selecting two green marbles is 12 out of 132.

We can repeat this process for two yellow marbles. We need to multiply eight twelfths by seven elevenths. Multiplying the numerators here gives us 56, and multiplying the denominators gives us 132. The probability of selecting two yellow marbles is 56 out of 132. Each of these two parts means that we have selected two marbles that are the same color. Selecting two green marbles or selecting two yellow marbles will give us what we want. We recall that when dealing with tree diagrams, the OR rule means add. So we need to add these two fractions.

When adding two fractions with the same denominator, we simply add the numerators. And 12 plus 56 is equal to 68. The probability of choosing two marbles that are the same color is 68 out of 132. As both the numerator and denominator are divisible by four, we can simplify this fraction. 68 divided by four is 17. And 132 divided by four is 33. The probability of choosing two marbles that are the same color written in its simplest form is 17 out of 33.

The third and final part of our question wants us to calculate the probability of choosing one marble of each color. This could happen by choosing a green marble followed by a yellow marble or a yellow marble followed by a green marble. To calculate the probability of green and then yellow, we need to multiply four twelfths by eight elevenths. Four multiplied by eight is 32. So this is equal to 32 out of 132. Selecting a yellow marble followed by a green marble also gives us 32 out of 132. This time, weโre multiplying eight twelfths by four elevenths. As either of these combinations gives us one marble of each color, we need to add our two fractions, giving us 64 out of 132. This fraction once again simplifies by dividing the numerator and denominator by four. The probability of selecting one marble of each color is 16 out of 33.

There is a quicker way we couldโve found this answer by using the answer from the second part of the question. We know that there are four possible combinations from this tree diagram: green green, green yellow, yellow green, and yellow yellow. These four probabilities must sum to one. And we know that the top and bottom probabilities sum to 17 out of 33. Subtracting this from one gives us 16 out of 33 as 33 minus 17 is equal to 16. This means that the other two options green yellow and yellow green must sum to 16 over 33.

We have now completed all three parts of this question. The values of ๐ด, ๐ต, ๐ถ, and ๐ท were four twelfths, three elevenths, eight elevenths, and seven elevenths. The probability of choosing two marbles of the same color was 17 out of 33. And the probability of choosing one marble of each color was 16 out of 33. The answers to the second and third part of the question are the complement of each other as they sum to one.