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
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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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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.
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Addition Rule of Probability
It simply refers to the likelihood of an event taking place whenever the occurrence of an event is uncertain. The probability of a single event can be calculated by dividing the number of successful trials of that event by the total number of trials.
Expected Value
When a large number of trials are performed for any random variable ‘X’, the predicted result is most likely the mean of all the outcomes for the random variable and it is known as expected value also known as expectation. The expected value, also known as the expectation, is denoted by: E(X).
Probability Distributions
Understanding probability is necessary to know the probability distributions. In statistics, probability is how the uncertainty of an event is measured. This event can be anything. The most common examples include tossing a coin, rolling a die, or choosing a card. Each of these events has multiple possibilities. Every such possibility is measured with the help of probability. To be more precise, the probability is used for calculating the occurrence of events that may or may not happen. Probability does not give sure results. Unless the probability of any event is 1, the different outcomes may or may not happen in real life, regardless of how less or how more their probability is.
Basic Probability
The simple definition of probability it is a chance of the occurrence of an event. It is defined in numerical form and the probability value is between 0 to 1. The probability value 0 indicates that there is no chance of that event occurring and the probability value 1 indicates that the event will occur. Sum of the probability value must be 1. The probability value is never a negative number. If it happens, then recheck the calculation.
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- A manufacturer has determined that a machine averages one faulty unit for every 500 it produces. What is the probability that an order of 300 units will have one or more faulty units?
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Step 1: Determine whether each probability is greater than or equal to 0 and less than or equal to 1. Step 2: Determine whether the sum of all of the probabilities equals 1. Step 3: If Steps 1 and ...
Statistics as Inductive Inference. Jan-Willem Romeijn, in Philosophy of Statistics, 2011. 7 Bayesian Statistics. The defining characteristic of Bayesian statistics is that probability assignments do not just range over data, but that they can also take statistical hypotheses as arguments. As will be seen in the following, Bayesian inference is naturally represented in terms of a non-ampliative ...
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 ...
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.
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.
Yes or No. (b) Give reason for your answer in 5 (a). Carry intermediate calculations to at least four decimal places before rounding to. 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 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? Explain. No, they are greater than or equal to 0, but do not sum to 1. v - Select your answer - Yes, they are greater than or equal to 0.
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…
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}.
The simple definition of probability it is a chance of the occurrence of an event. It is defined in numerical form and the probability value is between 0 to 1. The probability value 0 indicates that there is no chance of that event occurring and the probability value 1 indicates that the event will occur. Sum of the probability value must be 1.
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 ...
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 ...
A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: P(E1) = 0.05, P(E2) = 0.15, P(E3) = 0.40, and P(E4) = 0.35. Are these probability assignments valid? Explain. No, the subjective method requires that all probabilities be equal. No, the probabilities do not sum to 1.
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 (E)-0.10, P (E₂) 0.05, P (E)-0.35, and P (E)-0.45. Are these probability assignments valid?
The probability assignments mentioned in the question are valid. According to the rules of probability, the probability of an event, denoted as P(E), must be between 0 and 1, inclusive. In this case, all the probabilities mentioned for E1, E2, E3, and E4 fall within this range.
Math. Statistics and Probability. Statistics and Probability questions and answers. 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? Choose matching definition What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons? .678 b.
The sum of probabilities assigned to all four outcomes equals 1, meeting the requirement for a valid probability assignment. Explanation: (a) Yes, the given probability assignment is valid because the probabilities assigned to the four possible outcomes add up to 1, which is a requirement for a valid probability assignment. In this case, P(E1 ...
Are these probability assignments valid? Explain. O No, the subjective method requires that all probabilities be equal. Yes, 0 S P(E;) s 1 for all i and the probabilities' sum is less than 1. Yes, 0 = P(E;) < 1 for all i and the probabilities' sum is greater than 1. No, the probabilities do not sum to 1.
The probability assignments given by the decision maker are not valid because they do not sum to 1. The sum of probabilities for all outcomes in an experiment should always be equal to 1. In this case, the sum of probabilities is equal to 0.10 + 0.15 + 0.40 + 0.20 = 0.85, which is less than 1. Learn more about Probability Assignments here: