Unit 8 Probability And Sampling — Unit Plan

The more complex a situation is, the harder it can be to estimate the probability of a particular event happening. Well-designed simulations are a way to estimate a probability in a complex situation, especially when it would be difficult or impossible to determine the probability from reasoning alone.

To design a good simulation, we need to know something about the situation. For example, if we want to estimate the probability that it will rain every day for the next three days, we could look up the weather forecast for the next three days. Here is a table showing a weather forecast:

We can set up a simulation to estimate the probability of rain each day with three bags.

• In the first bag, we put 4 slips of paper that say “rain” and 6 that say “no rain.”
• In the second bag, we put 5 slips of paper that say “rain” and 5 that say “no rain.”
• In the third bag, we put 9 slips of paper that say “rain” and 1 that says “no rain.”

Then we can select 1 slip of paper from each bag and record whether or not the simulation predicts that there will be rain on all three days. If we repeat this experiment many times, we can estimate the probability that there will be rain on all three days by dividing the number of times all three slips say “rain” by the total number of times we perform the simulation.

Sometimes we need a systematic way to count the number of outcomes that are possible in a given situation. For example, suppose there are 3 people (A, B, and C) who want to run for the president of a club and 4 different people (1, 2, 3, and 4) who want to run for vice president of the club. We can use a tree, a table, or an ordered list to count how many different combinations are possible for a president to be paired with a vice president.

With a tree, we can start with a branch for each of the people who want to be president. Then for each possible president, we add a branch for each possible vice president, for a total of $3\boldcdot 4 = 12$ possible pairs. We can also start by counting vice presidents first and then adding a branch for each possible president, for a total of $4 \boldcdot 3 = 12$ possible pairs.

Tree diagram with three branches for the first choice, labeled “A,” “B”, and “C.” Choices “A”, “B”, and “C” each have four branches labeled with a different number from 1 through 4.

Tree diagram with four branches for the first choice, labeled 1, 2, 3, and 4. Choices 1, 2, 3, and 4 each have three branches, labeled with a different letter “A,” “B,” or “C.”  

A table can show the same result:

So does this ordered list:

A1, A2, A3, A4, B1, B2, B3, B4, C1, C2, C3, C4

Suppose we have two bags. One contains 1 star block and 4 moon blocks. The other contains 3 star blocks and 1 moon block.

If we select 1 block at random from each, what is the probability that we will get 2 star blocks or 2 moon blocks?

To answer this question, we can draw a tree diagram to see all of the possible outcomes.

A tree diagram. The first choice has 5 branches, representing the 5 blocks in the bag: one branch is labeled “star,” the other 4 are labeled “moon.” Each of these branches has 4 branches, representing the 4 blocks in the second bag. 3 branches are labeled “star” and one is labeled “moon.” The word “star” in the first choice, and the 3 “star” choices branching from it are highlighted gold. From the first choice, the word “moon” is highlighted blue on each of the four remaining branches. From each of those branches, the one choice of “moon” for each is also highlighted blue.

There are $5 \boldcdot 4 = 20$ possible outcomes. Of these, 3 of them are both stars, and 4 are both moons. So the probability of getting 2 star blocks or 2 moon blocks is $\frac{7}{20}$.

In general, if all outcomes in an experiment are equally likely, then the probability of an event is the fraction of outcomes in the sample space for which the event occurs.

Many real-world situations are difficult to repeat enough times to get an estimate for a probability. If we can find probabilities for parts of the situation, we may be able to simulate the situation using a process that is easier to repeat.

For example, if we know that each egg of a fish in a science experiment has a 13% chance of having a mutation, how many eggs do we need to collect to make sure we have 10 mutated eggs? If getting these eggs is difficult or expensive, it might be helpful to have an idea about how many eggs we need before trying to collect them.

We could simulate this situation by having a computer select random numbers between 1 and 100. If the number is between 1 and 13, it counts as a mutated egg. Any other number would represent a normal egg. This matches the 13% chance of each fish egg having a mutation.

We could continue asking the computer for random numbers until we get 10 numbers that are between 1 and 13. How many times we asked the computer for a random number would give us an estimate of the number of fish eggs we would need to collect.

To improve the estimate, this entire process should be repeated many times. Because computers can perform simulations quickly, we could simulate the situation 1,000 times or more.

Some populations have greater variability than others. For example, we would expect greater variability in the weights of dogs at a dog park than at a beagle meetup.

Dog park:

Mean weight: 12.8 kg       MAD: 2.3 kg

Beagle meetup:

Mean weight: 10.1 kg       MAD: 0.8 kg

The lower MAD indicates that there is less variability in the weights of the beagles. We would expect that the mean weight from a sample that is randomly selected from a group of beagles will provide a more accurate estimate of the mean weight of all the beagles than a sample of the same size from the dogs at the dog park.

In general, if samples from a population have similar sizes, a sample with less variability is more likely to have a mean that is close to the population mean.

Sometimes a data set consists of information that fits into specific categories. For example, we could survey students about whether they have a pet cat or dog. The categories for these data would be {neither, dog only, cat only, both}. Suppose we surveyed 10 students. Here is a table showing possible results:

In this sample, 3 of the students say they have both a dog and a cat. We can say that the proportion of these students who have a both a dog and a cat is $\frac{3}{10}$ or 0.3. If this sample is representative of all 720 students at the school, we can predict that about $\frac{3}{10}$ of 720, or about 216 students at the school, have both a dog and a cat.

In general, a proportion is a number from 0 to 1 that represents the fraction of the data that belongs to a given category.

Sometimes we want to compare two different populations. For example, is there a meaningful difference between the weights of pugs and beagles? Here are histograms showing the weights for a sample of dogs from each of these breeds:

A histogram for two different populations: On the horizontal axis, the numbers 6 through 11, in increments of zero point 5, are indicated. The label “pug weights in kilograms” is indicated for the numbers 6 through 8 and “beagle weights in kilograms” is indicated for the numbers 9 through 11. On the vertical axis, the numbers 0 through 8 are indicated. The data represented by the bars are as follows: Pug weights in kilograms: Weight from 6 up to 6 point 5, 5. Weight from 6 point 5 up to 7, 5. Weight from 7 up to 7 point 5, 7. Weight from 7 point 5 up to 8, 3. A triangle is located at 6 point 9 kilograms. Beagle weights in kilograms: Weight from 9 up to 9 point 5, 3. Weight from 9 point 5 up to 10, 3. Weight from 10 up to 10 point 5, 8. Weight from 10 point 5 up to 11, 6. A triangle is located at 10 point 1.  

The red triangles show the mean weight of each sample, 6.9 kg for the pugs and 10.1 kg for the beagles. The red lines show the weights that are within 1 MAD of the mean. We can think of these as typical weights for the breed. These typical weights do not overlap. In fact, the distance between the means is $10.1-6.9$, or 3.2 kg, over 6 times the larger MAD! So we can say there is a meaningful difference between the weights of pugs and beagles.

Is there a meaningful difference between the weights of male pugs and female pugs? Here are box plots showing the weights for a sample of male pugs and a sample of female pugs:

Two box plots labeled “male pug weights in kilograms” and “female pug weights in kilograms” are indicated. The numbers 4 through 8 point 5, in increments of zero point 5, are indicated. The five-number summary for the box plots are as follows: Male pug weights in kilograms: Minimum value, 6 point 4. Maximum value, 8 point 3. Q1, 7 point 2. Q2, 7 point 6. Q3, 7 point 9. Female pug weights in kilograms: Minimum value, 6 point 2. Maximum value, 8. Q1, 6 point 4. Q2, 6 point 9. Q3, 7 point 3.  

We can see that the medians are different, but the weights between the first and third quartiles overlap. Based on these samples, we would say there is not a meaningful difference between the weights of male pugs and female pugs.

In general, if the measures of center for two samples are at least two measures of variability apart, we say the difference in the measures of center is meaningful. Visually, this means the ranges of typical values do not overlap. If they are closer, then we don't consider the difference to be meaningful.