# 8.1.4: Estimating Probabilities Through Repeated Experiments

## Lesson

Let's do some experimenting.

Exercise (PageIndex{1}): Decimals on the Number Line

1. Locate and label these numbers on the number line.
1. (0.5)
2. (0.75)
3. (0.33)
4. (0.67)
5. (0.25)
2. Choose one of the numbers from the previous question. Describe a game in which that number represents your probability of winning. Exercise (PageIndex{2}): In the Long Run

Mai plays a game in which she only wins if she rolls a 1 or a 2 with a standard number cube.

1. List the outcomes in the sample space for rolling the number cube.
2. What is the probability Mai will win the game? Explain your reasoning.
3. If Mai is given the option to flip a coin and win if it comes up heads, is that a better option for her to win?
4. Begin by dragging the gray bar below the toolbar down the screen until you see the table in the top window and the graph in the bottom window. This applet displays a random number from 1 to 6, like a number cube. Mai won with the numbers 1 and 2, but you can choose any two numbers from 1 to 6. Record them in the boxes in the center of the applet.
Click the Roll button for 10 rolls and answer the questions.
5. What appears to be happening with the points on the graph?
1. After 10 rolls, what fraction of the total rolls were a win?
2. How close is this fraction to the probability that Mai will win?
6. Roll the number cube 10 more times. Record your results in the table and on the graph from earlier.
1. After 20 rolls, what fraction of the total rolls were a win?
2. How close is this fraction to the probability that Mai will win?

Exercise (PageIndex{3}): Due For A Win

1. For each situation, do you think the result is surprising or not? Is it possible? Be prepared to explain your reasoning.
1. You flip the coin once, and it lands heads up.
2. You flip the coin twice, and it lands heads up both times.
3. You flip the coin 100 times, and it lands heads up all 100 times.
2. If you flip the coin 100 times, how many times would you expect the coin to land heads up? Explain your reasoning.
3. If you flip the coin 100 times, what are some other results that would not be surprising?
4. You’ve flipped the coin 3 times, and it has come up heads once. The cumulative fraction of heads is currently (frac{1}{3}). If you flip the coin one more time, will it land heads up to make the cumulative fraction (frac{2}{4})?

### Summary

A probabilityfor an event represents the proportion of the time we expect that event to occur in the long run. For example, the probability of a coin landing heads up after a flip is (frac{1}{2}), which means that if we flip a coin many times, we expect that it will land heads up about half of the time.

Even though the probability tells us what we should expect if we flip a coin many times, that doesn't mean we are more likely to get heads if we just got three tails in a row. The chances of getting heads are the same every time we flip the coin, no matter what the outcome was for past flips.

### Glossary Entries

Definition: Chance Experiment

A chance experiment is something you can do over and over again, and you don’t know what will happen each time.

For example, each time you spin the spinner, it could land on red, yellow, blue, or green. Definition: Event

An event is a set of one or more outcomes in a chance experiment. For example, if we roll a number cube, there are six possible outcomes. Examples of events are “rolling a number less than 3,” “rolling an even number,” or “rolling a 5.”

Definition: Outcome

An outcome of a chance experiment is one of the things that can happen when you do the experiment. For example, the possible outcomes of tossing a coin are heads and tails.

Definition: Probability

The probability of an event is a number that tells how likely it is to happen. A probability of 1 means the event will always happen. A probability of 0 means the event will never happen.

For example, the probability of selecting a moon block at random from this bag is (frac{4}{5}). Definition: Random

Outcomes of a chance experiment are random if they are all equally likely to happen.

Definition: Sample Space

The sample space is the list of every possible outcome for a chance experiment.

For example, the sample space for tossing two coins is:

## Practice

Exercise (PageIndex{4})

A carnival game has 160 rubber ducks floating in a pool. The person playing the game takes out one duck and looks at it.

• If there’s a red mark on the bottom of the duck, the person wins a small prize.
• If there’s a blue mark on the bottom of the duck, the person wins a large prize.
• Many ducks do not have a mark.

After 50 people have played the game, only 3 of them have won a small prize, and none of them have won a large prize.

Estimate the number of the 160 ducks that you think have red marks on the bottom. Then estimate the number of ducks you think have blue marks. Explain your reasoning.

Exercise (PageIndex{5})

Lin wants to know if flipping a quarter really does have a probability of (frac{1}{2}) of landing heads up, so she flips a quarter 10 times. It lands heads up 3 times and tails up 7 times. Has she proven that the probability is not (frac{1}{2})? Explain your reasoning.

Exercise (PageIndex{6})

A spinner has four equal sections, with one letter from the word “MATH” in each section.

1. You spin the spinner 20 times. About how many times do you expect it will land on A?
2. You spin the spinner 80 times. About how many times do you expect it will land on something other than A? Explain your reasoning.

Exercise (PageIndex{7})

A spinner is spun 40 times for a game. Here is a graph showing the fraction of games that are wins under some conditions. Estimate the probability of a spin winning this game based on the graph.

Exercise (PageIndex{8})

Which event is more likely: rolling a standard number cube and getting an even number, or flipping a coin and having it land heads up?

(From Unit 8.1.2)

Exercise (PageIndex{9})

Noah will select a letter at random from the word “FLUTE.” Lin will select a letter at random from the word “CLARINET.”

Which person is more likely to pick the letter “E?” Explain your reasoning.

(From Unit 8.1.3)

## 8.1.4: Estimating Probabilities Through Repeated Experiments

Probability represents the expected likelihood of an event occurring for a single trial on an experiment.

• Regardless of previous tries, each coin flip should still be equally likely to land heads up as tails up.
• A basketball player who tends to make 75% of his free throw shots will probably make ¾ free throws he attempts, but there is no guarantee that he will make any individual shot, even if he has missed a few in a row.

You conduct a chance experiment many times and record the outcomes. How are these outcomes related to the probability of a certain event occurring?

• The fraction of times the event occurs after many repetitions should be fairly close to the expected probability of the event.

What is the probability of rolling a 2, 3, or 4 on a standard number cube?

You want to roll a 2, 3, or 4 on a standard number cube. If you roll 3 times and none of them result in a 2, 3, or 4, does the probability of getting one of those numbers change with the next roll?

The probability of getting the flu during flu season is ⅛. If a family has 8 people living in the same house, is it guaranteed that one of them will get the flu?

• No, it’s very possible that none of the people will get the flu, and also possible that more than 1 person will get the flu.

The probability of getting the flu during flu season is ⅛. If a country has 8 million people, about how many do you expect will get the flu? Does this number have to be exact?

• I can explain why it may be useful to gather data on a sample of a population.
• When I read or hear a statistical question, I can name the population of interest and give an example of a sample for that population.
• I remember that when a distribution is not symmetric, the median is a better estimate of a typical value than the mean.
• I can determine whether a sample is representative of a population by considering the shape, center, and spread of each of them.
• I know that some samples may represent the population better than others.

## PERT distribution

The PERT distribution is a continuous probability distribution that is defined by the minimum ( a ), most likely ( b ) and maximum ( c ) values, with the mean (or expected value) of:

The PERT distribution is actually a transformed four parameter beta distribution. The beta distribution have two parameters α and β defined on the interval [0,1], making it useful for modelling probabilities and random variables. The four parameter version of the beta distribution makes the interval [a,c] instead where a is the minimum and c is the maximum value.

You can see where this is going. By using a PERT distribution to map the estimations, we can approximate the probabilities of the estimations themselves!

This is to say, while we ask the estimator to provide the optimistic, pessimistic and most likely values for the estimations, the actual expected value could be different from the most likely value, and we can figure out how likely the tasks is going to be completed within a duration or a budget.

However this is really only for a single task. In a project, of course, we have many, many tasks and each might have different distribution parameters for a, b and c. If you’re the project manager, how would you figure out the overall estimations and probabilities?

Well, we can also try to do the maths but we’re programmers, not mathematicians (I’m saying this in a good way) so we look towards our old friend, Monte Carlo.

## Sampling with or without replacement

Sometimes we might want to repeat the same experience multiple times, like flipping a coin. When repeating experiments, a common assumption is that the outcome of one experiment does not have any influence on the outcome of the others. In order words, the experiments are independent. In real-life situations, we often select things from a larger population and analyze them. There are the following two ways to do this: sampling with replacement, sampling without replacement.

When we sequentially select with replacement, the outcomes can repeat and the experiments are independent.

In the example below, we sample two balls from a jar with two balls, one yellow and the second blue. If we select without replacement from the jar, the outcomes cannot repeat and the experiments are dependent. In all examples so far, order did matter. Sometimes the order does not matter, and we just care about the collection. Instead of working with tuples of possible outcomes, we deal with the set of possible outcomes.

For example, if we flip a coin twice, we can have the outcomes (head, tail) or (tail, head). When order does not matter, both outcomes build the event <(head, tail), (tail, head)>.

Let’s now look at another situation, where we select 2 cards out of a deck with 6 cards with replacement. When order matters, we have a uniform distribution, meaning that all possible tuples (1,1), (1,2), …, (6,6,) have the same probability of occurring 1/36.

However, when order does not matter, the distribution is not uniform anymore, as illustrated below. Let’s repeat the same experiment, this time without replacement. Now, the distribution is uniform, as shown below. Order and sampling methods can have a significant impact on the probabilities. ## 8.1.4: Estimating Probabilities Through Repeated Experiments

PROBABILITY/STATISTICS IN INHERITANCE

We know that when two people who are both heterozygous for a simple Mendelian autosomal gene alpha have a child, the probability that the child will show the dominant phenotype is 3/4. Let's ask a somewhat more complex question. If this couple has a total of four children, what is the probability that 3 of the 4 will show the dominant phenotype? To answer this, we will first derive the appropriate formula and then use it to calculate the numerical answer. The same formula allows us to understand the expected statistical distribution of the various possible phenotype patterns in four-child ( or any other size) families in a large population.

1. Review: How can some simple overall probabilities be calculated by combining the multiplication and addition "rules" we covered earlier?

What is the probability that all three children in a family will be the same gender?
P(all female)= 1/2 x 1/2 x 1/2 = 1/8
P(all male ) = 1/2 x 1/2 x 1/2 = 1/8
P(all one gender) = P(all female) + P(all male) = 1/8 + 1/8 = 1/4

What is the probability that a three-child family is two girls and one boy?
Each possible birth order has P=1/8. That is, P(G,G,B)=P(G,B,G)=P(B,G,G)=1/8.
So, P(2G,1B)= 3/8 and P(1G,2B)= 3/8.

This allows us to write the overall gender probability distribution for families of three children as follows:
1/8 will be three girls
3/8 will be two girls and one boy
3/8 will be one girl and two boys
1/8 will be three boys
Adding it all up, we have 1/8 + 3/8 + 3/8 + 1/8 = 1 (100%)

2. How can we understand and use "Pascal's triangle" and the "general rule for repeated trials of events with constant probabilities", formula 1 (page 161 in textbook)?

Consider the numerators of the fractions in the above three-child family gender equation:
1 , 3 , 3 , 1. These numbers are the coefficients in the expansion of the term (p + q) cubed. In general, the coefficients of any such binomial expansion < the term (p+q) raised to any power>give the "number of ways" that something can happen.

Figure 4.21, "Pascal's triangle", shows these coefficients for the expansion of (p + q) raised to any power up to 10. The numbers in any row can be used just as described above. For example, assume that over the next two decades you have 6 kids. There are 64 possible gender birth orders, with 20 of these resulting in you having three girls and three boys.

The terms p and q are the individual probabilities for a specific outcome from a single "event". For "gender" calculations, the probabilities p and q are equal, both = 1/2 (the equal probabilities of male and female births).

For "dominant : recessive phenotype" type of calculations, p and q will usually not be equal. For a simple Mendelian inheritance from two heterozygotic parents, p will = 3/4 (if AA and Aa give dominant phenotype) and q will = 1/4 (aa gives recessive phenotype).

Generalizing this, we get to the formula on page 161 in your text that is "the general rule for repeated trials of events with constant probabilities". The term (n!/s!t!) is the number of possible ways (orders) of getting a certain net outcome ( "a total of n with s of one and t of the other"). This number can either be calculated or taken directly from Pascal's triangle.

3. Sample problem: You and your mate are both heterozygous for some simple Mendelian gene alpha (i.e., each of you has genotype Aa, and both of you show the dominant phenotype) on chromosome #1. Over the next decade, you proceed to have four children. What is the probability that 3 of your children will show the dominant phenotype and one will show the recessive phenotype? What are the probabilities of the other possible outcomes?

If we were looking at thousands of such families, we know that the overall ratio of dominant to recessive phenotypes in the children would average out to 3:1, as shown by a simple Punnett square. But for one couple having four children, what's the probability, P(3D,1r)?

To calculate P(3D,1r), we use formula 1 for the case n=4, s=3, t=1, p=3/4, q=1/4.
P(3D,1r)= 4!/3! x (3/4)cubed x (1/4) = 4 x (27/64) x 1/4 = .42 ( 42% )

By also calculating the other four possibilities, we can construct a graph that shows the statistical distribution you would expect to see in a large population.

Problem S-4: "Heterozygous parents have three children".

Modify the sample problem above to do the calculations for you and your mate (both Aa) having THREE children. Do the calculations for the probabilities that all three, two, one, or none of the three children will show the dominant phenotype from gene alpha. Construct the graph and compare the result with the "four children" graph done in class.

## Abstract

Accurately estimating driving styles is crucial to designing useful driver assistance systems and vehicle control systems for autonomous driving that match how people drive. This paper presents a novel way to identify driving style not in terms of the durations or frequencies of individual maneuver states, but rather the transition patterns between them to see how they are interrelated. Driving behavior in highway traffic was categorized into 12 maneuver states, based on which 144 (12 × 12) maneuver transition probabilities were obtained. A conditional likelihood maximization method was employed to extract typical maneuver transition patterns that could represent driving style strategies, from the 144 probabilities. Random forest algorithm was adopted to classify driving styles using the selected features. Results showed that transitions concerning five maneuver states – free driving, approaching, near following, constrained left and right lane changes – could be used to classify driving style reliably. Comparisons with traditional methods were presented and discussed in detail to show that transition probabilities between maneuvers were better at predicting driving style than traditional maneuver frequencies in behavioral analysis.

## 8.1.4: Estimating Probabilities Through Repeated Experiments

1. Among $A$, $B$, and $C$, only $A$ occurs: $A-B-C=A-(B cup C)$.
2. At least one of the events $A$, $B$, or $C$ occurs: $A cup B cup C$.
3. $A$ or $C$ occurs, but not $B$: $(A cup C)-B$.
4. At most two of the events $A$, $B$, or $C$ occur: $(A cap B cap C)^c=A^c cup B^c cup C^c$.

The Venn diagrams are shown in Figure 1.19. Fig.1.19 - Venn diagrams for solved problem 1.
1. We toss a coin until we see two consecutive tails. We record the total number of coin tosses.
2. A bag contains $4$ balls: one is red, one is blue, one is white, and one is green. We choose two distinct balls and record their color in order.
3. A customer arrives at a bank and waits in the line. We observe $T$, which is the total time (in hours) that the customer waits in the line. The bank has a strict policy that no customer waits more than $20$ minutes under any circumstances.

Remember that the sample space is the set of all possible outcomes. Usually, when you have a random experiment, there are different ways to define the sample space $S$ depending on what you observe as the outcome. In this problem, for each experiment it is stated what outcomes we observe in order to help you write down the sample space $S$.

1. We toss a coin until we see two consecutive tails. We record the total number of coin tosses: Here, the total number of coin tosses is a natural number larger than or equal to $2$. The sample space is $S=<2,3,4,cdots>.$
2. A bag contains $4$ balls: one is red, one is blue, one is white, and one is green. We choose two distinct balls and record their color in order: The sample space can be written as $S=<(R,B),(B,R),(R,W),(W,R),(R,G),(G,R),$ $(B,W),(W,B),(B,G),(G,B),(W,G),(G,W)>.$
3. A customer arrives at a bank and waits in the line. We observe $T$. In theory $T$ can be any real number between $and$frac<1><3>=20$minutes. Thus,$S=ig[0,frac<1><3>ig]=ig | 0 leq x leq frac<1><3>ig>.$•$A cup B cup C=S$, •$P(A)=frac<1><2>$, •$P(B)=frac<2><3>$, •$P(A cup B)=frac<5><6>$. 1. Find$P(A cap B)$. 2. Do$A$,$B$, and$C$form a partition of$S$? 3. Find$Pig(C-(A cup B)ig)$. 4. If$P(C cap (A cup B))=frac<5><12>$, find$P(C)$. As before, it is always useful to draw a Venn diagram however, here we provide the solution without using a Venn diagram. Using the inclusion-exclusion principle, we have$P(A cup B)=P(A)+P(B)-P(A cap B).$Thus, $C-(A cup B)=igg(C cup (Acup B)igg)-(Acup B)hspace<30pt>=S-(A cup B) extrm<(since $A cup B cup C=S$)>=(A cup B)^c$. Thus As we saw before, the sample space$S$has$36$elements. 1. This is an example of a continuous probability model. Write down the sample space$S$. 2. Check that the statement in the manual makes sense by finding$P(T geq 0)$and$lim_ P(T geq t)$. 3. Also check that if$t_1 e^<-frac<5>>$=P(T geq t_2)$ (since $f(x)=e^<(x)>$ is an increasing function). Here we have two events, $A$ is the event that $T geq t_1$ and $B$ is the event that $T geq t_2$. That is, $A =[t_1,infty), B=[t_2, infty).$ Since $B$ is a subset of $A$, $B subset A$, we must have $P(B)leq P(A)$, thus $P(A)=P(T geq t_1)geq P(T geq t_2)=P(B).$
4. The probability that the product breaks down within three years of the purchase time is $P(T Problem I first saw this question in a math contest many years ago: You get a stick and break it randomly into three pieces. What is the probability that you can make a triangle using the three pieces? You can assume the break points are chosen completely at random, i.e. if the length of the original stick is$1$unit, and$x, y, z$are the lengths of the three pieces, then$(x,y,z)$are uniformly chosen from the set$<(x,y,z) in mathbb^3 | x+y+z=1, x,y,z geq 0>.$This is again a problem on a continuous probability space. The basic idea is pretty simple. First, we need to identify the sample space$S$. In this case the sample space is going to be a two-dimensional set. Second, we need to identify the set$A$that contains the favorable outcomes (the set of$(x,y,z)$in$S$that form a triangle). And finally, since the space is uniform, we will divide area of set$A$by the area of$S$to obtain$P(A)$. First, we need to find the sets$S$and$A$. This is basically a geometry problem. The two sets,$S$and$A$, are shown in Figure 1.20. Fig.1.20 - The sample space and set$A$for Problem 6. Note that in$mathbb^3$,$x+y+z=1$represents a plane that goes through the points$(1,0,0), (0,1,0), (0,0,1)$. To find the sample space$S$, note that$S=<(x,y,z) in mathbb^3 | x+y+z=1, x,y,z geq 0>$, thus$S$is the part of the plane that is shown in Figure 1.20. To find the set$A$, note that we need$(x,y,z)$to satisfy the triangle inequality$x+y > z,y+z > x,x+z > y.$Note that since$x+y+z=1$, we can equivalently write the three equations as$x < frac<1><2>,y < frac<1><2>,z < frac<1><2>.$Thus, we conclude that the set$A$is the area shown in Figure 20. In particular, we note that the set$S$consists of four triangles with equal areas. Therefore, its area is four times the area of$A$, and we have$P(A)=frac < extrmA> < extrmS>=frac<1><4>.\$

## Difference between experimental probability and theoretical probability You can argue the same thing using a die. We will though use a coin to help you see the difference.

In theoretical probability, we say that "each outcome is equally likely " without the actual experiment.ਏor instance, without ਏlipping a coin, you know that the outcome could either be heads or tails.  If the coin is not altered, we argue that each outcome ( heads or tails ) is equally likely. In other words, we are saying that in theory or ( supposition, conjecture, speculation, assumption, educated guess) the probability to get heads is 50% or the probability to get tails in 50%. Since you did not actually flip the coin, you are making an assumption based on logic.

The logic is that there are 2 possible outcomes and since you are choosing 1 of the 2 outcomes, the probability is 1/2 or 50%. This is theoretical probability or guessing probability or probability based on assumption.

In experimental probability,  we want to take the guess work out of the picture, by doing the experiment to see how many times heads or teals will come up. If you flip a coin 1000 times, you might realize that it landed on heads only 400 times. In this case, the probability to get heads is only 40%.

Your experiment may not even show tails until after the 4th flip and yet in the end you ended up with more tails than heads.

If you repeat the experiment another day, you may find a completely different result. May be this time the coin landed on tails 400 times or 300 times.

As you can see, experimental probability is based more on facts, data collected, experiment or research!

## Solved Examples on Experimental Probability

Example 1: The following table shows the recording of the outcomes on throwing a 6-sided die 100 times.

Outcome Frequency
1 14
2 18
3 24
4 17
5 13
6 14

Find the experimental probability of: a) Rolling a four b) Rolling a number less than four c) Rolling a 2 or 5

Solution:
Experimental probability is calculated by the formula: Number of times an event occurs/Total number of trials
a) Rolling a 4: 17/100 = 0.17