/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Determine whether the informatio... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 44

Determine whether the information shown is consistent with a probability distribution. If not, say why. \(P(A)=.2 ; P(B)=.4 ; P(A \cap B)=.3\)

Short Answer

Expert verified
The given information is consistent with a probability distribution. We used the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) to calculate the probability of the union of A and B, and then found a complementary event C such that the sum of probabilities of all events is equal to 1. The final probability distribution is: \(P(A)=0.2, P(B)=0.4, P(A \cap B)=0.3, P(A \cup B)=0.3, P(C)=0.7\).

Step by step solution

01

Check Non-negativity

All the given probabilities are non-negative, as they are all positive decimal values: \(P(A)=0.2 \ge 0, P(B)=0.4 \ge 0\) and \(P(A \cap B)=0.3 \ge 0\)
02

Check if sum of probabilities equal to 1

We can't directly sum the given probabilities and check if it equals 1, as we are given the probabilities of two events and their intersection, not the probabilities of mutually exclusive events. However, we can rearrange these values to determine whether the information is consistent. We can use the mathematical formula of \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) to find the probability of the union of events A and B. Then we can determine if there is a complementary event, say event C, such that the probabilities of all events add up to 1.
03

Calculate Probability of Union of A and B

Using the given probabilities, let's calculate the probability of the union of events A and B: \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.2 + 0.4 - 0.3 = 0.3\)
04

Determine Consistency

We found that \(P(A \cup B) = 0.3\). Now if there exists an event C (complementary event), such that \(P(C)=1-P(A \cup B)\), then it would be consistent with a probability distribution. Let's calculate the probability of event C: \(P(C) = 1 - P(A \cup B) = 1 - 0.3 = 0.7\) Since we found a probability for event C such that the sum of the probabilities of all events is equal to 1, the given information is consistent with a probability distribution. The final probability distribution is: \(P(A)=0.2, P(B)=0.4, P(A \cap B)=0.3, P(A \cup B)=0.3, P(C)=0.7\)

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Key Concepts

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

Non-negativity
In probability theory, one of the foundational properties of a probability distribution is non-negativity. This means that the probability of any event cannot be negative. It is always a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In the given exercise, we checked this property by evaluating the probabilities of events A, B, and their intersection (A 鈭� B).
  • Probability of A: \(P(A) = 0.2\), and clearly \(0.2 \geq 0\).
  • Probability of B: \(P(B) = 0.4\), where \(0.4 \geq 0\).
  • Probability of both A and B occurring together: \(P(A \cap B) = 0.3\), ensuring \(0.3 \geq 0\).
All given probabilities meet the non-negativity requirement, meaning none of the values fall below zero. Meeting this condition is crucial before determining if the set of probabilities forms a proper distribution.
Probability of Union
The probability of the union of two events, denoted as \(P(A \cup B)\), represents the probability that at least one of the events A or B occurs. To calculate this, we use the formula:\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]This formula accounts for any overlap between the events A and B, ensuring that their intersection is not counted twice. In the provided exercise, the probability of the union was calculated as follows:
  • \(P(A \cup B) = 0.2 + 0.4 - 0.3 = 0.3\)
This calculation shows that there is a 0.3 probability that either event A, event B, or both occur. Calculating the union is essential when examining multiple events, as it helps to understand their combined probability effectively.
Complementary Events
Complementary events refer to two mutually exclusive events whose probabilities sum up to 1. In probability terms, if an event A has a probability \(P(A)\), the complementary event \(A'\) (not A) has a probability \(1 - P(A)\). Together, these two represent the entire sample space.During the solution, once the probability of the union \(P(A \cup B)\) was calculated as 0.3, identifying a complementary event became possible. This hypothetical event C complements the union of A and B. Its probability is given by:\[P(C) = 1 - P(A \cup B) = 1 - 0.3 = 0.7\]This calculation shows that the remaining probability, not covered by A or B, is captured by event C. Identifying complementary events is necessary to ensure all possibilities within a probability distribution are considered, confirming the total probability is 1.

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Most popular questions from this chapter

Based on the following table, which shows the performance of a selection of 100 stocks after one year. (Take S to be the set of all stocks represented in the table.) $$ \begin{array}{|r|c|c|c|c|} \hline & \multicolumn{3}{|c|} {\text { Companies }} & \\ \cline { 2 - 4 } & \begin{array}{c} \text { Pharmaceutical } \\ \boldsymbol{P} \end{array} & \begin{array}{c} \text { Electronic } \\ \boldsymbol{E} \end{array} & \begin{array}{c} \text { Internet } \\ \boldsymbol{I} \end{array} & \text { Total } \\ \hline \begin{array}{r} \text { Increased } \\ \boldsymbol{V} \end{array} & 10 & 5 & 15 & 30 \\ \hline \begin{array}{r} \text { Unchanged }^{*} \\ \boldsymbol{N} \end{array} & 30 & 0 & 10 & 40 \\ \hline \begin{array}{r} \text { Decreased } \\ \boldsymbol{D} \end{array} & 10 & 5 & 15 & 30 \\ \hline \text { Total } & 50 & 10 & 40 & 100 \\ \hline \end{array} $$ If a stock stayed within \(20 \%\) of its original value, it is classified as "unchanged." Use symbols to describe the event that a stock's value increased but it was not an Internet stock. How many elements are in this event?

If \(B \subseteq A\) and \(P(B) \neq 0\), why is \(P(A \mid B)=1 ?\)

Describe the sample space \(S\) of the experiment and list the elements of the given event. (Assume that the coins are distinguishable and that what is observed are the faces or numbers that face up.) HINT [See Examples 1-3.] A letter is chosen at random from those in the word Mozart; the letter is a vowel.

Use counting arguments from the preceding chapter. Gummy Bears A bag contains six gummy bears. Noel picks four at random. How many possible outcomes are there? If one of the gummy bears is raspberry, how many of these outcomes include the raspberry gummy bear?

Based on the following table, which shows the performance of a selection of 100 stocks after one year. (Take S to be the set of all stocks represented in the table.) $$ \begin{array}{|r|c|c|c|c|} \hline & \multicolumn{3}{|c|} {\text { Companies }} & \\ \cline { 2 - 4 } & \begin{array}{c} \text { Pharmaceutical } \\ \boldsymbol{P} \end{array} & \begin{array}{c} \text { Electronic } \\ \boldsymbol{E} \end{array} & \begin{array}{c} \text { Internet } \\ \boldsymbol{I} \end{array} & \text { Total } \\ \hline \begin{array}{r} \text { Increased } \\ \boldsymbol{V} \end{array} & 10 & 5 & 15 & 30 \\ \hline \begin{array}{r} \text { Unchanged }^{*} \\ \boldsymbol{N} \end{array} & 30 & 0 & 10 & 40 \\ \hline \begin{array}{r} \text { Decreased } \\ \boldsymbol{D} \end{array} & 10 & 5 & 15 & 30 \\ \hline \text { Total } & 50 & 10 & 40 & 100 \\ \hline \end{array} $$ If a stock stayed within \(20 \%\) of its original value, it is classified as "unchanged." \(\nabla\) Calculate \(\frac{n(D \cap I)}{n(D)}\). What does the answer represent?
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