22 Multistage random experiments 2

Slide 0

Today we’re going to talk about mathematics again. Welcome to this episode.

Slide 1

We have already talked about multistage random experiments. Today we will work a little more intensively with them and how probabilities are determined for these experiments.

Of course, one possibility is to determine probabilities empirically and for some questions, that may be the only possibility; we’re thinking of the thumbtack toss. However, in some cases, you can also make theoretical observations. Then the idea is to determine the favorable cases and to derive a theoretical probability.

Today we’re going to deal especially with the second case, but not only that, and at any rate with how you can suitably represent the results of a multistage random experiment.

Slide 2

First, let’s say a word about empiricism. You’re already familiar with this example from the previous episode.

We surveyed a class to find out whether the students like the subject English and/or the subject math. Here are the results. Why shouldn’t we interpret the relative numbers gained empirically as the probability?

If we were to pick one person from the class without looking, then the probability that this person likes English AND math is exactly ½ and the probability that this person likes math in general is 0.68 or 68 percent. Here we don’t have an opportunity to obtain the results in another way besides the poll.

Slide 3

With some multistage random experiments, however, you can also determine probabilities from a theoretical perspective. Let’s look at an example, specifically, flipping a coin twice.

You can come up with heads or tails on the first flip just like on the second flip. The probability of flipping “heads” both times is – as we all know – 25 percent or 1/4.

In this case, we do not necessary have to conduct the experiment, but we can also fill in the table from a theoretical point of view. Each time in one fourth of the cases, you should come up with one of the four possible orders when you repeat the experiment many times. And clearly you can interpret this as the probability of the respective event.

I intentionally chose an experiment that results in such boring numbers. They are boring, but so simple that you can understand the correlation immediately. The “heads” and “tails” events occur with a probability of 1/2 for a single flip, and each combination of two flips has a probability of 1/4 because there are exactly four different possibilities when – as a reminder – you take the order into account.

Slide 4

The theoretical view works exactly the same way in this example.

We roll a die twice and want to know whether at least one “6” is rolled. This experiment has four possible outcomes:

• You roll a “6” both times.
• You do not roll a “6” either time.
• It works only on the first roll.
• It works only on the second roll.

Slide 5

Naturally, you can really roll a die twice and at any rate you should do that enough times.

But in this case as well, we don’t necessarily have to actually conduct the random experiment. You can also fill in the table from a theoretical point of view. The question is, which theoretical point of view? Let’s look systematically at this.

Slide 6

On the first roll, there is one possibility of rolling a “6” and five possibilities of rolling a number other than “6”.

On the second roll, there is one possibility of rolling a “6” and five possibilities of rolling a number other than “6”.

To determine the probabilities theoretically, we can obviously apply the counting principle we know so well.

Slide 7

Let’s look at the situation in a tree diagram. There are a total of 36 different combinations of two natural numbers between 1 and 6. Among them, there is one possibility of rolling a 6 two times in succession. There are five possibilities of rolling first a 6 and then another number as well as five possibilities of rolling one of the numbers from 1 to 5 first and then a 6. Lastly, there are 5x5=25 possibilities of not getting a 6 in either of the two rolls.

 

Slide 8

We can enter these absolute numbers in a 2x2 table, which shows the results again very clearly. And once again, the total at the lower right consists of all 36 possibilities.

Slide 9

If we determine the relative frequencies from these numbers, we easily find the theoretical probabilities in this two-stage random experiment.

You see that this representation is very clear and allows you to easily check the values.

Slide 10

You’re already familiar with this example too.

We roll a die three times and want to know whether at least one “6” is rolled. Here are the theoretical results.

• You roll a “6” every time.
• You do not roll a “6” any of the times.
• It works only on the first roll.
• It works only on the second roll.
• It works only on the third roll.
• It works on the first and second rolls.
• It works on the first and third rolls.
• It works on the second and third rolls.

Slide 11

Can we assign probabilities here as well? Let’s try to get a handle on the situation using the general counting principle again.

Slide 12

Here we have arranged the situation in a tree diagram. As we have done previously, we can label the branches with the absolute values. For each branch, that is 1 for rolling a “6” or 5 for rolling another number. At the end we add the numbers up. Of 216 possible results, there is exactly one in which “6” comes up three times. There are 125 possibilities that “6” doesn’t come up at all in three rolls. There are 3 times 5 equals 15 possibilities in which a “6” is rolled twice, and 3 times 25 equals 75 possibilities for only one “6” and two other numbers.

Each of the possibilities can be entered, and we simply need three stages.

 This time, the theoretically possible different cases add up to 6³ = 216.

One more thing: It’s quite obvious that we cannot use a 2x2 table to represent this example. This table really is suitable only for two-stage random experiments.

Slide 13

These are the probabilities which here of course take into account the order of the rolls. The probability is 1/216 for rolling a “6” three times and 5/216 that another number comes up the third time after rolling two sixes. If you’re interested only in whether a “6” is rolled twice, then you can calculate 5/216 + 5/216 + 5/216 = 15/216.

Slide 14

Do you notice anything? Multiplication proves to be a useful cultural technique here. Multiplication of the corresponding numbers on the branches leads to the result on the nodes determined in this way. 1/6 • 1/6 • 1/6 = 1/216, 1/6 • 5/6 • 5/6 = 25/216, etc. We’ll come back to this in the next example.

Slide 15

Let’s take another look at a known situation under the new heading of multistage random experiments.

An urn contains two red balls and three blue balls. We draw a ball twice with replacement. Then there are a total of 25 possible results that can be easily determined by counting.

Slide 16

In this case, it makes sense to set

p(RR) = 4/25, p(BB) = 9/25, and p(RB) = 12/25

You can see this by counting. But also in this case, it works again using multiplication.

It’s quite clear that this new way of looking at things does not come with any surprises.

Slide 17

We have seen the following: In the multistage random experiments considered, the probabilities can be determined via the general counting principle. But ultimately, we just multiply in each case. And we can derive a rule from this, path rule no. 1:

We consider a multistage random experiment, represent it in a tree diagram, and enter the individual probabilities on the respective branches.

Then the probability of an experiment outcome, that is, of an elementary event, is the product of the probabilities along the corresponding branches.

Slide 18

Check for yourself and multiply. It clearly works.

p(blue, blue) = 3/5 • 3/5 = 9/25

p(red, red)    = 2/5 • 2/5 = 4/25

p(blue, red)   = 3/5 • 2/5 = 6/25

p(red, blue)   = 2/5 • 3/5 = 6/25

Slide 19

And if the order doesn’t matter and you consolidate, then clearly p(one red and one blue) = 6/25 + 6/25 = 12/25. The probabilities are added.

Slide 20

You can formulate the result as path rule no. 2.

We consider a multistage random experiment, represent it in a tree diagram, and enter the probability of an experiment outcome, that is, of an elementary event, at the end of the path. Then the probability of any random event corresponds to the sum of all probabilities of paths leading to this event.

Slide 21

Is this difficult to understand? No, it’s just not that easy to mathematically formulate the fact so that it’s really foolproof.

Let’s look at another example.

We roll a die twice and add up the number of spots. How high is the probability that the sum is at least “9”?

Slide 22

A tree diagram helps represent the situation. First, we enter the six possible results of the first roll, seen here as red nodes. Six paths lead from each of the results, but not all paths lead to a sum of 9 or greater. We draw only these paths, count how many there are, and multiply this number with 1/6 • 1/6 = 1/36.

Both path rules are applied in the process.

Slide 23

Do the path rules also work for drawing balls without replacement? But of course.

Let’s assume that an urn contains two red balls and three blue balls. We draw one ball from the urn, do not put it back, and draw another ball. You see the situation in the tree diagrams, and it is easy to determine the possibilities and thus the probabilities on the branches.

Slide 24

If we multiply, we clearly come up with these numbers. Drawing two red balls should occur in 2/20 or 1/10 of the cases. All other combinations are assigned the probability 6/20, thus 3/10.

Slide 25

Here as well, you can add if you’re not interested in the order. The probability is p(one red and one blue) = 6/20 + 6/20 = 12/20 = 3/5.

Slide 26

Because it fits in so well here, let’s briefly review the term Laplace experiment.

Rolling a die is a Laplace experiment. The results 1, 2, 3, 4, 5, and 6 occur with an equal probability of 1/6.

Rolling a die twice is a Laplace experiment (if you pay attention to the order). The results (1,1), (1,2), (1,3), …, (3,1), (3,2), …,(3,6), (4,1), … (6,5), (6,6) occur with an equal probability of 1/36.

In contrast, determining the sum of spots when rolling a die twice is not a Laplace experiment. The sum of “2” occurs only if a “1” is rolled twice. For the sum of “7,” there are the possibilities “1+6,” “2+5,” “3+4,” “4+3,” “5+2,” and “6+1,” and that’s clearly more possibilities.

Slide 27

What have we done in this episode? Well, we have looked at multistage random experiments and considered what the probability of a particular result looks like.

For some experiments, we can use the general counting principle to determine a theoretical frequency for the various results. Then we can assign probabilities accordingly.

It seems to make sense to represent the results in a tree diagram for all experiments. This representation allows us to make the experiment’s structure clear.

For two-stage random experiments only, a 2x2 table is also helpful and can be the basis for determining probabilities.

Slide 28

That’s all for today. Many thanks for being here. I look forward to seeing you next time.