### Video Transcript

A box contains seven green balls and five red balls. If a ball is drawn at random and replaced and then another ball is drawn, find the probability that the first one is red and the second one is green.

Let’s begin by just summarizing the information in the question. This box contains seven green and five red balls. So in total, there are 12, that’s seven plus five, balls in the box. We’re told that a ball is drawn at random from these 12. So that means that every ball in the box has an equal chance of being chosen. And then, the really key part of the information given in the question is that the ball that we’ve chosen is replaced. It’s put back into the box before the second selection is made. We’re then asked to find the probability that the first ball is red and the second is green, which we can do by using a tree diagram. This is a type of sample space which enables us to represent all the possible outcomes of these two events and their associative probabilities.

We begin by listing the possible outcomes for the color of the first ball. It can be either red or green. We then list the possible outcomes for the color of the second ball. And in fact, they are exactly the same. Regardless of whether the first ball was red or green, the second ball can still be either red or green. So we’ve got the structure for our tree diagram. And next, we need to consider the probabilities that go on each branch.

Let’s consider the probability that the first ball is red. Well, remember, we’re told that there are five red balls in the bag. And there’re 12 balls in total. So if this ball is chosen at random, then the probability that it is a red ball will be five out of 12. In the case of the first ball being green, well, there’re seven green balls in the bag out of the total of 12. So the probability that the first ball is green is seven out of 12 or seven twelfths. Notice that the sum of these two probabilities, five twelfths and seven twelfths, is one. And that’s no coincidence. This will always be the case for each set of branches on a tree diagram.

Next, we consider the probabilities for the color of the second ball. And what’s really key here is that the first ball was replaced in the bag. So what this means is that the setup when we’re choosing the second ball is identical to when we chose the first ball. Absolutely nothing has changed. The probability that the second ball is red then, regardless of what happened before, is the same as it was for the first ball. It’s five twelfths. And the probability that the second ball is green, regardless of what happened for the first ball, is the same as the probability that the first ball was green. It’s still seven twelfths.

So we now have our completed tree diagram. And we need to use it to find the probability that the first ball is red and the second one is green. That’s this outcome here. Well, to find this probability, we recall that we need to multiply together each of the probabilities on the branches that we need to take through our tree diagram. So that’s the probability of five twelfths that the first ball is red and then the probability of seven twelfths that the second ball is green. This is an application of the “and” rule of probability.

So we have that the probability that the first ball is red and the second is green is five twelfths multiplied by seven twelfths. Multiplying the numerators of these fractions together, five multiplied by seven is 35. And multiplying the denominators together, 12 multiplied by 12 is 144. This probability can’t be simplified any further. And there’s no need to convert it to a decimal.

So by drawing a tree diagram to represent all the possible outcomes in this situation, we found that the probability that the first ball is red and the second is green is 35 over 144.