Statistics Made Easy
How to Determine if a Probability Distribution is Valid
A probability distribution tells us the probability that a random variable takes on certain values.
In order for a probability distribution to be valid, it must meet two requirements:
1. Each probability must be between 0 and 1.
2. The sum of the probabilities must add up to 1.
If both of these requirements are met, then the probability distribution is valid.
The following examples show how to check if different probability distributions are valid.
Example 1: Goals Scored in a Soccer Game
The following probability distribution shows the probability of a certain soccer team scoring a certain number of goals in a game:
Let’s check if this probability distribution meets the two requirements to be valid:
We can see that each individual probability is between 0 and 1.
We can see that the sum of the probabilities adds up to 1:
Sum = .18 + .34 + .35 + .11 + .02 = 1
Both requirements are met so this probability distribution is valid .
Example 2: Sales Made in a Month
The following probability distribution shows the probability that a given salesman will make a certain number of sales in the upcoming month:
We can see that the sum of the probabilities does not add up to 1:
Sum = .44 + .31 + .39 + .06 = 1.2
Both requirements are not met so this probability distribution is not valid .
Example 3: Number of Battery Failures
The following probability distribution tells us the probability that a given vehicle experiences a certain number of battery failures during a 10-year span:
We can see that each individual probability is not between 0 and 1.
The last probability in the table is a negative value.
We can see that the sum of the probabilities does add up to 1:
Sum = .24 + .57 + .22 – .03 = 1
Additional Resources
The following tutorials provide additional information about probability distributions:
How to Find the Mean of a Probability Distribution How to Find the Variance of a Probability Distribution How to Find the Standard Deviation of a Probability Distribution
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2.4 - how to assign probability to events.
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:
- the personal opinion approach
- the relative frequency approach
- the classical approach
On this page, we'll take a look at each approach.
The Personal Opinion Approach Section
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 sometime this decade."
Example 2-4 Section
At which end of the probability scale would you put the probability that:
- one day you will die?
- you can swim around the world in 30 hours?
- you will win the lottery someday?
- a randomly selected student will get an A in this course?
- you will get an A in this course?
The Relative Frequency Approach Section
The relative frequency approach involves taking the follow three steps in order to determine P ( A ), the probability of an event A :
- Perform an experiment a large number of times, n , say.
- Count the number of times the event A of interest occurs, call the number N ( A ), say.
- 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 2-5 Section
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?
I think you all might instinctively reply \(\dfrac{1}{2}\). 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:
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 2-6 Section
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?
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?
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?
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?
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?
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 Section
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:
\(P(A)=\dfrac{N(A)}{N(\mathbf{S})}\)
where \(N(A)\) is the number of elements in the event \(A\), and \(N(\mathbf{S})\) is the number of elements in the sample space \(\mathbf{S}\). Let's take a look at an example.
Example 2-7 Section
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:
- \(A= \{x: x \text{ is a }2, 3,\text{ or }7\}\)
- \(B = \{x: x\text{ is 2H, 3D, 8S, or KC}\}\)
- What is the probability that a 2, 3, or 7 is drawn?
- What is the probability that the card is a 2 of hearts, 3 of diamonds, 8 of spades or king of clubs?
- 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?
- What is \(P(A\cap B)\)?
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Precalculus
Course: precalculus > unit 8.
- Constructing a probability distribution for random variable
Valid discrete probability distribution examples
- Graph probability distributions
- Probability with discrete random variable example
- Probability with discrete random variables
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Video transcript
SOLUTION: A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: P(E1) = .10, P(E2) = .15, P(E3) = .40, and P(E4) = .20. Are these probabil
Snapsolve any problem by taking a picture. Try it in the Numerade app?
COMMENTS
Let's check if this probability distribution meets the two requirements to be valid: 1. Each probability must be between 0 and 1. We can see that each individual probability is between 0 and 1. 2. The sum of the probabilities must add up to 1. We can see that the sum of the probabilities adds up to 1: Sum = .18 + .34 + .35 + .11 + .02 = 1.
Step 2: Determine whether the sum of all of the probabilities equals 1. The sum of the probabilities (or the sum of the entries in the second row) in the table is: {eq}0.6+0.2+0.1+0.05+0.05=1 {/eq ...
The relative frequency approach involves taking the follow three steps in order to determine P ( A ), the probability of an event A: Perform an experiment a large number of times, n, say. Count the number of times the event A of interest occurs, call the number N ( A ), say. Then, the probability of event A equals: P ( A) = N ( A) n.
Are these probability assignments valid? Explain. A.) No, the subjective method requires that all probabilities be equal. B.) Yes, 0 ≤ P(Ei) ≤ 1 for all i and the probabilities' sum is less than 1. C.) No, the probabilities do not sum to 1. D.) Yes, 0 ≤ P(Ei) ≤ 1 for all i and the probabilities' sum is greater than 1.
Microsoft Word - FA011108.doc. 1. Assigning Probabilities. The key to assigning probabilities is knowing all of your possible outcomes and knowing two rules: • All possible outcomes must total 1 or 100% (Where have we talked about 100% being important) • A probability must take a value 0 ≤ P(A) ≤ 1 (or 0% to 100%) 1) Probabilities can ...
This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. Question: A decision maker subjectively assigned the following probabilities to the all the four possible outcomes of an activity: P (E1) = 0.35, P (E2) = 0.12, P (E3) = 0.44, and P (E4) = 0.20.
A valid probability model must account for all possible outcomes of the experiment. If not all possibilities are displayed, it's reasonable that the sum of the probabilities won't add up to 1 (or 100%). This omission could be why the total probability in the example does not reach 100%, suggesting an incomplete model.
Probability Law of Total Probability Bayes' Rule 6.5 Example A decision maker subjectively assigned the following probabilities to the four possible outcomes of an experiment: P(E 1) =0:1 2 15 3 4 4 2 Are these probability assignments legitimate? Explain. Solution. Probability II Union, Intersection, and Logical Relationships among Events ...
A basic probability assignment is a mapping m from the set of all subsets of a universal finite nonempty set X to the interval [0, 1], but it need not meet all the probability measure requirements. Its main features are ∑ S⊂X m ( S) = 1 and m (∅) = 0.
That is, are there other possible probability assignments that lead to a probabilty of 1/4 of hitting the target on any toss, or is yours the only valid assignment? Carefully prove your answer. 2.17. A spinner as shown in the figure selects a number from the set {1, 2, 3, 4}.
to check if the probability assignments assigned are valid, you have to check that the probabilities meet two requirements. The first requirement is that each probability is between zero and one 0.1 point 15.4 and point to are all between zero and one and thus meet the…
Are these probability assignments valid? Explain. Answer by Alex.33(110) (Show Source): You can put this solution on YOUR website! If there are JUST four outcomes of the experiment, it's invalid. Because .1+.15+.4+.2=.85, which is not equal to 1. ... it will be valid JUST if the sum of the possibilities of other outcomes is .15, but otherwise ...
This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. Question: A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: P (E₁) = 0.10, P (E2) = 0.15, P (E3) = 0.40, and P (EA) = 0.20. Are these probability assignments valid?
MCQ. A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: P(E1) = 0.10, P(E2) = 0.15, P(E3) = 0.40, and P(E4) = 0.20. Are these probability assignments valid? Select: Yes, they are greater than or equal to 0 Yes, they are greater than or equal to 0 and sum to 1
{/eq} Are these probability assignment valid? Explain. Probability. The probability is a possible chance of an event. The value of probability must fall between 0 and 1. If there are 5 events, then the sum of all probabilities must be equal to 1. ... Learn the probability definition in math, and how to solve probability examples in math, and ...
In this item, we are to explain whether the probability assignments are valid or not. To do so, we recall the two basic requirements for assigning probabilities. First, each of the probabilities must be between 0 0 0 and 1 1 1. Second, the sum of the assigned probabilities must be equal to 1.0 1.0 1.0. We are then given the following assigned ...
Statistics and Probability. Statistics and Probability questions and answers. HW Problems 1) A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: P (E1) .10, PIE2) 15. PEB) 40, and P (64) 20. Are these probability assignments valid?
Are these proba bility assignments valid? Explain. Math. Statistics. A decision maker subjectively assigned the following probabilities to the four outcomes ofan experiment: P(E1) = .10, P(E2) = .15, P(E3) = .40, and P(E4) = .20. Are these proba bility assignments valid? Explain. ... Compute the probability of finding oil: The probability of ...
A decision maker subjectedly assigned the following probabilities to the four outcomes of an experiment: P (E1)=.10, P (E2)=.15, P (E3)=.40, and P (E4)=.20. Are these probability assignments valid? Explain. Holt Mcdougal Larson Pre-algebra: Student Edition 2012. 1st Edition.
You'll get a detailed solution from a subject matter expert that helps you learn core concepts. Question: a) A decision maker subjectively assigned the following probabilities to the four outcomes of a random experiment: P (E1) = 0.05, P (E2) = 0.10, P (E3) = 0.40 and P (E4) = 0 .20. i) Write S. ii) Are these probability assignments valid?
Are these probability assignments valid? Explain. No, the probabilities do not sum to 1. An experiment consists of four outcomes with P (E 1) = 0.2, P (E 2) = 0.3, and P (E 3) = 0.4. The probability of outcome E 4 is. 0.100. An experiment consists of three steps. There are four possible results on the first step, three possible results on the ...
A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: P(E)-0.10, P(E₂) 0.05, P(E)-0.35, and P(E)-0.45. Are these probability assignments valid? Explain. No, the probabilities do not sum to 1. No, the subjective method requires that all probaboties be equal.
Solution (a) These are not valid probability assignments The probabilities s …. Strom Construction made a bid on two contracts. The owner identified the possible outcomes and subjectively assigned the following probabilities, Experimental Outcome Obtain Contract 1 Obtain Contract 2 Probability 1 Yes Yes 0.14 2 Yes NO 0.13 3 No Yes 0.28 4 No ...