Video: Determining the Probability of Intersection of Two Dependent Events

A bag contains 8 red balls and 8 black balls. If two balls are drawn without replacement, what is the probability of getting one red ball and one black ball?

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Video Transcript

A bag contains eight red balls and eight black balls. If two balls are drawn without replacement, what is the probability of getting one red ball and one black ball?

A key piece of the wording within the question is that the two balls are drawn without replacement, which means that the first ball is not put back into the bag before the second ball is chosen. Effectively, the two balls are selected at the same time. This has an impact on the probabilities for the second ball. And we must consider this when we’re answering the question.

We can use a tree diagram to help us represent all of the outcomes for the colors of the two balls. Our first set of branches is for the color of the first ball, which can be either red or black, represented by 𝑅 and 𝐵. Our second set of branches is for the color of the second ball, which can also be either red or black. There are, therefore, four possibilities for the colors of the two balls. They could both be red, the first is red and the second is black, the first is black and the second is red, or both balls are black.

We’re interested in the probability of getting one red ball and one black ball. And as it hasn’t been specified that the red ball must come before the black ball, this means that we’re interested in the two middle possibilities on our tree diagram, red black or black red.

We need to consider the probabilities on the tree diagram itself so that we can work out the probabilities of these two events. The probabilities on the first set of branches are relatively straightforward. To begin with, there are eight red and eight black balls in the bag. So, in total, there are 16 balls in this bag. There are eight red, so the probability of choosing a red ball is eight over 16. And there are eight black, so the probability of choosing a black ball is also eight over 16. We can simplify these probabilities to one-half. But we won’t do this until we’ve worked out the probabilities on the second set of branches of our tree diagram.

Now, key point here, there were 16 balls in the bag to begin with. But remember, the balls are drawn without replacement. So, by the time we come to choose the second ball, there are now only 15 balls left in the bag because one has been removed. So, the probabilities for the second set of branches are all going to have denominators of 15. The numerators, though, will vary depending on what has already happened. If the first ball selected was red, then there will now be seven red balls left in the bag. So, the probability that the second ball will be red if the first was red is seven out of 15. If the first ball was red, there is still eight black balls left in the bag. So, the probability that the second ball will be black if the first ball was red is eight out of 15.

On the lower set of branches, if a black ball was selected first of all, then there are still eight red balls left in the bag. So, the probability that the second ball is red given that the first was black is eight over 15. There were only seven black balls left in the bag. So, the probability that the second ball is black if the first was black is seven out of 15.

We can see that the probability of the second ball being red is different depending on the color of the first ball. Now we need to work out the probabilities for these middle branches on the tree diagram. That’s the probability of getting a red and then a black and then the probability of getting a black and then a red.

To work these out, we multiply along the branches of the tree diagram. So, the probability of getting a red and then a black is found by multiplying eight over 16 by eight over 15. Now we will simplify that probability of eight over 16 to one-half at this point because it makes our multiplication much more straightforward. We can also cancel a factor of two from the two in the denominator of the first fraction and the eight in the numerator of the second, giving one over one multiplied by four over 15. This product just simplifies to four over 15. So, the product of getting a red and then a black ball is four fifteenths.

Now, by symmetry, the probability of getting a black and then a red will be the same as the probability we’ve just worked out for getting a red and then a black. Because at the start of the question, the situation is identical for red and black. There are eight red and eight black balls in the bag. But we can demonstrate this longhand anyway. The probability of getting a black and then a red ball is eight over 16, which we simplify to one-half, multiplied by eight over 15. As before, we cancel a factor of two, and the product is equal to four fifteenths.

Now, at this point, we have our two individual probabilities. But we need to combine them together. We note then that we’re looking for the probability of getting a red and then a black or the probability of getting a black and then a red. The OR rule in probability tells us that if two events 𝐴 and 𝐵 are mutually exclusive, meaning they don’t overlap or can’t happen at the same time, then the probability of 𝐴 or 𝐵 occurring is equal to the sum of their individual probabilities, the probability of 𝐴 plus the probability of 𝐵.

The events of getting a red and then a black or a black and then a red are mutually exclusive. It’s one or the other. These two things can’t happen at the same time. So, to find the probability of getting red then black or black then red, we add four fifteenths to four fifteenths. These two fractions have the same denominator. So, we just add their numerators, giving eight fifteenths. We’ve found then that the probability of getting one red ball and one black ball in either order is eight fifteenths.

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