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Chapter 2: Problem 73

Use the axioms of probability to show the following: (a) For any event \(E, P\left(E^{\prime}\right)=1-P(E)\). (b) \(P(\varnothing)=0\) (c) If \(A\) is contained in \(B,\) then \(P(A) \leq P(B)\).

Short Answer

Expert verified
(a) Proven using complement rule, (b) via probability axiom, (c) by subset principle.

Step by step solution

01

Understand the Complement Rule

According to the Complement Rule of probability, for any event \(E\), the probability of the complement of \(E\) (denoted as \(E'\)) is equal to one minus the probability of \(E\). This can be expressed as \(P(E') = 1 - P(E)\). This is derived from the axiom that the probability of the entire sample space is 1, \(P(S) = 1\), and since \(E \cup E' = S\) and \(E \cap E' = \varnothing\), it follows that \(P(E) + P(E') = 1\), leading to \(P(E') = 1 - P(E)\).
02

Prove Probability of Empty Set

According to the probability axioms, the probability of the empty set \(\varnothing\) is 0. The empty set represents an impossible event, which means it has no chance of occurring. Formalizing this, for any event \(A\), \(A \cap \varnothing = \varnothing\) and \(A \cup \varnothing = A\). Therefore, no additional probability is added by the empty set, i.e., \(P(A \cup \varnothing) = P(A)\). Since there's no change, \(P(\varnothing) = 0\).
03

Prove Probability of Subsets

To demonstrate that if \(A\) is contained in \(B\), then \(P(A) \leq P(B)\), we rely on the Axiom of Probability which states that if an event \(A\) is a subset of another event \(B\), then \(B\) contains all outcomes of \(A\). Thus, \(B\) has at least the same or more probability of occurring than \(A\), since \(A \subseteq B\) implies \(A = A \cap B\). Therefore, by the addition law, \(P(B) = P(A) + P(B - A)\) (where \(B - A\) represents elements of \(B\) not in \(A\)) which implies \(P(A) \leq P(B)\).

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Key Concepts

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

Complement Rule
The complement rule is a key concept in the field of probability. It revolves around the likelihood of an event not occurring. An event and its complement are all the possibilities within a sample space that do not belong to that event. As expressed mathematically, for an event \(E\), the probability of the complement \(E'\) is given by:\[ P(E') = 1 - P(E) \]This means, if you already know how likely event \(E\) is to happen, you can easily find out the chance of it not happening. This is because together, the event and its complement cover the entire sample space, which always has a total probability of 1. - **Sample Space**: All possible outcomes combined - **Event E**: The occurrence we are focusing on- **Complement E'**: All outcomes not in event E

The complement rule is extremely useful when probabilities are complex, but figuring out the opposite is simpler.
Empty Set Probability
In probability, the concept of the empty set, represented as \(\varnothing\), is fundamental. Although it might sound unfamiliar, it is just another way to describe an impossible event鈥攕omething that has no chance of happening.The empty set has a probability of zero:\[ P(\varnothing) = 0 \]This arises from one of the axioms of probability that explains that no elements exist in the empty set to contribute any probability mass. Therefore, when no outcomes are possible, the probability is naturally zero. Here's how these properties are expressed:- **Intersection with Any Event**: \(A \cap \varnothing = \varnothing\)- **Union with Any Event**: \(A \cup \varnothing = A\)These properties highlight that the empty set doesn't change the probability of any real event \(A\) when it is involved in combinations.
Subset Probability
Understanding the probability relationship between a subset and its superset is also crucial. If you have two events, \(A\) and \(B\), and knowing that \(A\) is a subset of \(B\) implies the outcomes within \(A\) are also found in \(B\).This can be mathematically expressed as:\[ P(A) \leq P(B) \]The reasoning here is simple: since \(B\) includes every outcome of \(A\), plus possibly more, the probability of \(B\) occurring is at least as great as that of \(A\). Through the probability axioms and addition rules, this becomes clear:- **Subset Inclusion**: \(A \subseteq B\) implies all outcomes of \(A\) are in \(B\)- **Addition Law**: \(P(B) = P(A) + P(B - A)\)Using this concept helps us compare the likelihood of events, especially when one event is an expanded version of another.

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Most popular questions from this chapter

A machine tool is idle \(15 \%\) of the time. You request immediate use of the tool on five different occasions during the year. Assume that your requests represent independent events. (a) What is the probability that the tool is idle at the time of all of your requests? (b) What is the probability that the machine is idle at the time of exactly four of your requests? (c) What is the probability that the tool is idle at the time of at least three of your requests?

A computer system uses passwords that contain exactly eight characters, and each character is one of the 26 lowercase letters \((a-z)\) or 26 uppercase letters \((A-Z)\) or 10 integers \((0-9)\). Let \(\Omega\) denote the set of all possible passwords. Suppose that all passwords in \(\Omega\) are equally likely. Determine the probability for each of the following: (a) Password contains all lowercase letters given that it contains only letters (b) Password contains at least 1 uppercase letter given that it contains only letters (c) Password contains only even numbers given that is contains all numbers

A computer system uses passwords that contain exactly eight characters, and each character is one of the 26 lowercase letters \((a-z)\) or 26 uppercase letters \((A-Z)\) or 10 integers \((0-9)\). Let \(\Omega\) denote the set of all possible password, and let \(A\) and \(B\) denote the events that consist of passwords with only letters or only integers, respectively. Suppose that all passwords in \(\Omega\) are equally likely. Determine the following robabilities: (a) \(P\left(A \mid B^{\prime}\right)\) (b) \(P\left(A^{\prime} \cap B\right)\) (c) \(P\) (password contains exactly 2 integers given that it contains at least 1 integer)

Computer keyboard failures are due to faulty electrical connects ( \(12 \%\) ) or mechanical defects \((88 \%) .\) Mechanical defects are related to loose keys \((27 \%)\) or improper assembly \((73 \%)\). Electrical connect defects are caused by defective wires \((35 \%)\) improper connections \((13 \%),\) or poorly welded wires \((52 \%)\) (a) Find the probability that a failure is due to loose keys. (b) Find the probability that a failure is due to improperly connected or poorly welded wires.

An article in the British Medical Journal ["Comparison of treatment of renal calculi by operative surgery, percutaneous nephrolithotomy, and extracorporeal shock wave lithotripsy" (1986, Vol. 82, pp. \(879-892\) ) ] provided the following discussion of success rates in kidney stone removals. Open surgery had a success rate of \(78 \%(273 / 350)\) and a newer method, percutaneous nephrolithotomy (PN), had a success rate of \(83 \%(289 / 350)\). This newer method looked better, but the results changed when stone diameter was considered. For stones with diameters less than 2 centimeters, \(93 \%(81 / 87)\) of cases of open surgery were successful compared with only \(83 \%(234 / 270)\) of cases of PN. For stones greater than or equal to 2 centimeters, the success rates were \(73 \%(192 / 263)\) and \(69 \%(55 / 80)\) for open surgery and PN, respectively. Open surgery is better for both stone sizes, but less successful in total. In \(1951,\) E. H. Simpson pointed out this apparent contradiction (known as Simpson's paradox), and the hazard still persists today. Explain how open surgery can be better for both stone sizes but worse in total.
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