How To Write A Conclusion For An Assignment Of Probability

Printer-friendly version

We know that probability is a number between 0 and 1.  How does an event get assigned a particular probability value?  Well, there are three ways of doing so:

  1. the personal opinion approach
  2. the relative frequency approach
  3. the classical approach

On this page, we'll take a look at each approach.

The Personal Opinion Approach

This approach is the simplest in practice, but therefore it also the least reliable.  You might think of it as the "whatever it is to you" approach.  Here are some examples:

  • "I think there is an 80% chance of rain today."
  • "I think there is a 50% chance that the world's oil reserves will be depleted by the year 2100."
  • "I think there is a 1% chance that the men's basketball team will end up in the Final Four some time this decade."

Example

At which end of the probability scale would you put the probability that:

  1. one day you will die? 
  2. you can swim around the world in 30 hours? 
  3. you will win the lottery some day? 
  4. a randomly selected student will get an A in this course? 
  5. you will get an A in this course?

Solution. I think we'd all agree that the probability that you will die one day is 1.  On the other hand, the probability that you can swim around the world in 30 hours is nearly 0, as is the probability that you will win the lottery some day.  I am going to say that the probability that a randomly selected student will get an A in this course is a probability in the 0.20 to 0.30 range.  I'll leave it to you think about the probability that you will get an A in this course.

The Relative Frequency Approach

The relative frequency approach involves taking the follow three steps in order to determine P(A), the probability of an event A:

  1. Perform an experiment a large number of times, n, say.
  2. Count the number of times the event A of interest occurs, call the number N(A), say.
  3. Then, the probability of event A equals:

  \(P(A)=\dfrac{N(A)}{n}\)

The relative frequency approach is useful when the classical approach that is described next can't be used.

Example

When you toss a fair coin with one side designated as a "head" and the other side designated as a "tail", what is the probability of getting a head?

Solution. I think you all might instinctively reply ½.  Of course, right?  Well, there are three people who once felt compelled to determine the probability of getting a head using the relative frequency approach:  

Coin Tosser

n, the number of tosses made

N(H), the number of heads tossedP(H)
Count Buffon4,0402,0480.5069
Karl Pearson24,00012,0120.5005
John Kerrich10,0005,0670.5067

As you can see, the relative frequency approach yields a pretty good approximation to the 0.50 probability that we would all expect of a fair coin. Perhaps this example also illustrates the large number of times an experiment has to be conducted in order to get reliable results when using the relative frequency approach.

By the way, Count Buffon (1707-1788) was a French naturalist and mathematician who often pondered interesting probability problems.  His most famous question:

Suppose we have a floor made of parallel strips of wood, each the same width, and we drop a needle onto the floor. What is the probability that the needle will lie across a line between two strips?

came to be known as Buffon's needle problem.  Karl Pearson (1857-1936) effectively established the field of mathematical statistics. And, once you hear John Kerrich's story, you might understand why he, of all people, carried out such a mind-numbing experiment.  He was an English mathematician who was lecturing at the University of Copenhagen when World War II broke out.  He was arrested by the Germans and spent the war interned in a prison camp in Denmark. To help pass the time he performed a number of probability experiments, such as this coin-tossing one.

Example

Some trees in a forest were showing signs of disease. A random sample of 200 trees of various sizes was examined yielding the following results:

What is the probability that one tree selected at random is large?

Solution. There are 68 large trees out of 200 total trees, so the relative frequency approach would tell us that the probability that a tree selected at random is large is 68/200 = 0.34.

What is the probability that one tree selected at random is diseased?

Solution. There are 37 diseased trees out of 200 total trees, so the relative frequency approach would tell us that the probability that a tree selected at random is diseased is 37/200 = 0.185.

What is the probability that one tree selected at random is both small and diseased?

Solution. There are 8 small, diseased trees out of 200 total trees, so the relative frequency approach would tell us that the probability that a tree selected at random is small and diseased is 8/200 = 0.04.

What is the probability that one tree selected at random is either small or disease-free?

Solution. There are 121 trees (35 + 46 + 24 + 8 + 8) out of 200 total trees that are either small or disease-free, so the relative frequency approach would tell us that the probability that a tree selected at random is either small or disease-free is 121/200 = 0.605.

What is the probability that one tree selected at random from the population of medium trees is doubtful of disease?

Solution. There are 92 medium trees in the sample.  Of those 92 medium trees, 32 have been identified as being doubtful of disease. Therefore, the relative frequency approach would tell us that the probability that a medium tree selected at random is doubtful of disease is 32/92 = 0.348.

The Classical Approach

The classical approach is the method that we will investigate quite extensively in the next lesson.  As long as the outcomes in the sample space are equally likely (!!!), the probability of event A is:

\(P(A)=\dfrac{N(A)}{N(S)}\)

where N(A) is the number of elements in the event A, and N(S) is the number of elements in the sample space S.  Let's take a look at an example.

Example

Suppose you draw one card at random from a standard deck of 52 cards.  Recall that a standard deck of cards contains 13 face values (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, and King) in 4 different suits (Clubs, Diamonds, Hearts, and Spades) for a total of 52 cards.  Assume the cards were manufactured to ensure that each outcome is equally likely with a probability of 1/52.  Let A be the event that the card drawn is a 2, 3, or 7.  Let B be the event that the card is a 2 of hearts (H), 3 of diamonds (D), 8 of spades (S) or king of clubs (C). That is:

  • A = {x: x is a 2, 3, or 7}
  • B = {x: x is 2H, 3D, 8S, or KC}

Then:

  1. What is the probability that a 2, 3, or 7 is drawn?
  2. What is the probability that the card is a 2 of hearts, 3 of diamonds, 8 of spades or king of clubs?
  3. What is the probability that the card is either a 2, 3, or 7 or a 2 of hearts, 3 of diamonds, 8 of spades or king of clubs?
  4. What is P(AB)?

Solution. The following viewlet illustrates: (refresh your browser if necessary to see the viewlet)

Stats: Introduction to Probability


Sample Spaces

A sample space is the set of all possible outcomes. However, some sample spaces are better than others.

Consider the experiment of flipping two coins. It is possible to get 0 heads, 1 head, or 2 heads. Thus, the sample space could be {0, 1, 2}. Another way to look at it is flip { HH, HT, TH, TT }. The second way is better because each event is as equally likely to occur as any other.

When writing the sample space, it is highly desirable to have events which are equally likely.

Another example is rolling two dice. The sums are { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }. However, each of these aren't equally likely. The only way to get a sum 2 is to roll a 1 on both dice, but you can get a sum of 4 by rolling a 1-3, 2-2, or 3-1. The following table illustrates a better sample space for the sum obtain when rolling two dice.

First DieSecond Die
123456
1234567
2345678
3456789
45678910
567891011
6789101112

Classical Probability

The above table lends itself to describing data another way -- using a probability distribution. Let's consider the frequency distribution for the above sums.

SumFrequencyRelative Frequency
211/36
322/36
433/36
544/36
655/36
766/36
855/36
944/36
1033/36
1122/36
1211/36

If just the first and last columns were written, we would have a probability distribution. The relative frequency of a frequency distribution is the probability of the event occurring. This is only true, however, if the events are equally likely.

This gives us the formula for classical probability. The probability of an event occurring is the number in the event divided by the number in the sample space. Again, this is only true when the events are equally likely. A classical probability is the relative frequency of each event in the sample space when each event is equally likely.

P(E) = n(E) / n(S)

Empirical Probability

Empirical probability is based on observation. The empirical probability of an event is the relative frequency of a frequency distribution based upon observation.

P(E) = f / n

Probability Rules

There are two rules which are very important.

All probabilities are between 0 and 1 inclusive

0 <= P(E) <= 1

The sum of all the probabilities in the sample space is 1

There are some other rules which are also important.

The probability of an event which cannot occur is 0.

The probability of any event which is not in the sample space is zero.

The probability of an event which must occur is 1.

The probability of the sample space is 1.

The probability of an event not occurring is one minus the probability of it occurring.

P(E') = 1 - P(E)

Continue and learn more about the rules of probability.


Table of Contents

0 comments

Leave a Reply

Your email address will not be published. Required fields are marked *