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Greg made up another question for a small quiz. He assigns the probabilities \(P(A)=0.6, P(B)=0.7, P(A | B)=0.1\) and asks for the probability \(P(A \text { or } B\) ). What is wrong with the probability assignments?

Short Answer

Expert verified
The probability assignments are wrong because \(P(A \text{ or } B) = 1.23\), which is impossible as probabilities cannot exceed 1.

Step by step solution

01

Verify Probability Range

Check that all given probabilities are within the valid range of 0 to 1. We have the probabilities: \(P(A)=0.6\), \(P(B)=0.7\), and \(P(A | B)=0.1\). All of these are within the valid range, so there are no issues regarding the rang of the values.
02

Calculate \(P(A ext{ and } B)\) Using Conditional Probability Formula

Use the formula for conditional probability, \(P(A | B) = \frac{P(A ext{ and } B)}{P(B)}\), to find \(P(A ext{ and } B)\). Rearrange the formula to get \(P(A ext{ and } B) = P(A | B) imes P(B)\). Substituting the known values, \(P(A ext{ and } B) = 0.1 imes 0.7 = 0.07\).
03

Calculate \(P(A ext{ or } B)\) Using the Addition Rule

The formula for the probability of the union of two events is \(P(A ext{ or } B) = P(A) + P(B) - P(A ext{ and } B)\). Substitute the known values: \(P(A ext{ or } B) = 0.6 + 0.7 - 0.07 = 1.3 - 0.07 = 1.23\).
04

Check Validity of \(P(A ext{ or } B)\)

The result for \(P(A ext{ or } B)\) is 1.23, which is greater than 1. This is not valid since probabilities cannot exceed 1. This indicates an inconsistency in the probability assignments.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
To understand conditional probability, start by recognizing it as the likelihood of an event occurring, given that another event has already happened. For instance, with events A and B, the conditional probability of A given B, denoted as \(P(A|B)\), is calculated using the formula: \[ P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \] In our problem, Greg assigned \(P(A|B) = 0.1\), which means that the probability of event A occurring is 0.1, given that event B has occurred. This formula is essential because it tells us how A is related to B when B has already happened. By applying this, you understand the dependency between two events, and you can calculate other probability aspects like the intersection of events.
Addition Rule in Probability
The addition rule helps calculate the probability of the occurrence of at least one of two events, often referred to as 'either-or' scenarios. This rule is formally stated as:\[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \] Here, \(P(A \text{ or } B)\) represents the probability that either event A or event B, or both, will happen.Subtraction of \(P(A \text{ and } B)\) is necessary to prevent double-counting the outcomes where both events may occur simultaneously. In Greg's exercise, applying the rule resulted in a probability greater than 1, showing a mistake in the assigned probabilities, as probabilities cannot exceed 1.Always remember to adjust overlaps in events using this addition rule to ensure accuracy in probability calculations.
Probability Range
In probability theory, every probability value must fall within a specific range, which is 0 to 1. These boundaries reflect the certainty ranges from impossibility (0) to absolute certainty (1).In this exercise, all initial probabilities \(P(A)=0.6\), \(P(B)=0.7\), and \(P(A | B)=0.1\) fall within this valid range. However, when calculating \(P(A \text{ or } B)\) using the addition rule, the result was 1.23, exceeding the valid maximum of 1. This provides a clear indication that the initial assignments are incorrect or inconsistent. Ensure every calculated probability respects this range, as anything outside this indicates errors, demanding recalibration of the given data.

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Most popular questions from this chapter

Involve a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: \(2,3,4,5,6,7,8,9,10,\) Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four \(10 \mathrm{s},\) etc., down to four \(2 \mathrm{s}\) in each deck. You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(3 \text { on } 1 \text { st card and } 10 \text { on } 2 \text { nd })\) (c) Find \(P(10 \text { on } 1 \text { st card and } 3 \text { on } 2 \text { nd })\) (d) Find the probability of drawing a 10 and a 3 in either order.

You roll two fair dice, a green one and a red one. (a) What is the probability of getting a sum of \(7 ?\) (b) What is the probability of getting a sum of \(11 ?\) (c) What is the probability of getting a sum of 7 or \(11 ?\) Are these outcomes mutually exclusive?

(a) Draw a tree diagram to display all the possible head-tail sequences that can occur when you flip a coin three times. (b) How many sequences contain exactly two heads? (c) Probability Extension Assuming the sequences are all equally likely, what is the probability that you will get exactly two heads when you toss a coin three times?

Probability Estimate: Wiggle Your Ears Can you wiggle your ears? Use the students in your statistics class (or a group of friends) to estimate the percentage of people who can wiggle their ears. How can your result be thought of as an estimate for the probability that a person chosen at random can wiggle his or her ears? Comment: National statistics indicate that about \(13 \%\) of Americans can wiggle their ears (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York).

Critical Thinking Suppose two events \(A\) and \(B\) are mutually exclusive, with \(P(A) \neq 0\) and \(P(B) \neq 0 .\) By working through the following steps, you'll see why two mutually exclusive events are not independent. (a) For mutually exclusive events, can event \(A\) occur if event \(B\) has occurred? What is the value of \(P(A | B) ?\) (b) Using the information from part (a), can you conclude that events \(A\) and \(B\) are not independent if they are mutually exclusive? Explain.

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