Video Transcript
A bag contained three red marbles, two yellow marbles, and six blue marbles. You take a marble at random from the bag and record its color. Then, without replacing the first marble, you take a second marble from the bag and record its color. There are three parts to this question. Given that the first marble you take is red, what is the probability that the second marble you take is also red? What is the probability that the second marble you take is red regardless of the color of the first marble you take? What is the probability that you take at least one red marble?
These questions involve conditional probability. And we recall that the conditional probability of event š“ is the probability of event š“ given that event šµ occurs and is written as shown. There are lots of ways of answering this type of problem, but in this question, we will use tree diagrams.
We are told in the question that we have three red, two yellow, and six blue marbles in a bag. A marble is taken at random from the bag and its color recorded. This marble could be red, yellow, or blue. As there are three red marbles and 11 marbles in total, the probability the first marble is red is three out of 11, or three elevenths. In the same way, the probability that the first marble is yellow is two elevenths, and the probability that the first marble is blue is six elevenths.
This marble is not replaced, and a second marble is taken from the bag. Once again, the second marble could be any one of the three colors. If we consider the top part of our tree diagram, we know that the first marble selected was red. This means that there are two red marbles remaining out of a total of 10. And the probability that the second marble is also red is two out of 10 or two-tenths.
There are still two yellow marbles left in the bag. So the probability the second marble is yellow is two-tenths. And there are six blue marbles left in the bag. So the probability the second marble is blue is six-tenths. Note that we could simplify all three of these fractions, but we will leave them as they are at present.
Letās now consider what happens if the first marble selected was yellow. There would still be three red marbles in the bag. So the probability the second marble selected is red would be three-tenths. As one yellow marble has been removed, there is only one remaining, and the probability the second marble is yellow is one-tenth. Once again, the probability that the second marble is blue is six-tenths.
Finally, letās consider what happens if the first marble selected was blue. There will still be three red marbles in the bag. So the probability the second marble is red is three-tenths. The two yellow marbles are also still in the bag, so the probability the second marble is yellow is two-tenths. As one blue marble has been removed, the probability that the second marble is blue given that the first one is blue is five-tenths.
At this stage, it is worth noting that the sum of the probabilities for each set of branches must equal one. For example, three elevenths plus two elevenths plus six elevenths is equal to eleven elevenths, which equals one.
Before going back to the specifics of this question, it is also worth noting that there are nine possible combinations of two marbles. We could have selected two reds, a red followed by a yellow, a red followed by a blue; alternatively, a yellow followed by a red, two yellows, or yellow followed by a blue; and finally, a blue followed by a red, a blue followed by yellow, or two blue marbles.
In order to calculate the probability of these nine combinations, we multiply along the branches. For example, to calculate the probability of selecting two red marbles, we multiply three elevenths by two-tenths. Multiplying the numerators and multiplying the denominators, we see that this is equal to six out of 110. Once again, we could simplify this fraction, but we will leave it as it is for now.
Repeating this process for red, yellow and red, blue, we have six out of 110 and 18 out of 110. The probabilities of the other six combinations are as shown. Once again, it is worth noting that all nine of these probabilities must sum to one. And in any question should we require more than one of these, we simply add up the results we need.
Letās now return to the specifics of this question. The first part of the question said the following. Given that the first marble you take is red, what is the probability that the second marble you take is also red? This is an example of conditional probability. We want to calculate the probability that the second marble is red given that the first marble was red. We can read this directly from the tree diagram. If the first marble is red, the probability that the second marble is also red is two-tenths.
As this is the final answer, it is worth simplifying. We can divide the numerator and denominator by two, giving us one-fifth. The answer to the first part of our question is one-fifth.
Letās now consider the second part of the question. What is the probability that the second marble is red regardless of the color of the first marble you take? In this question, we want the second marble to be red, but we donāt care what color the first marble is. This can happen in one of three ways. We can select a red then another red, a yellow then a red, or a blue then a red. We need to find the sum of these three probabilities. We therefore need to add six out of 110, another six out of 110, and 18 out of 110. As the denominators are the same, we simply add the numerators, giving us 30 out of 110.
This time, we can simplify our fraction by dividing the numerator and denominator by 10, giving us three out of 11 or three elevenths. The probability that the second marble is red regardless of the color of the first marble is three elevenths.
The final part of our question said the following. What is the probability you take at least one red marble? There are five different ways this can occur. We could select two red marbles, a red followed by yellow, a red followed by a blue, a yellow followed by a red, or a blue followed by a red. As with the second part of the question, we will need to find the sum of these probabilities.
We have six, six, 18, six, and 18 out of 110, respectively. This time the numerators sum to 54. So our probability is equal to 54 out of 110. And as both the numerator and denominator are divisible by two, this simplifies to 27 over 55. The probability that at least one red marble is selected is 27 out of 55.
We now have answers to the three parts of the question. They are one-fifth, three elevenths, and 27 out of 55.
