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Use the information that, for events \(\mathrm{A}\) and \(\mathrm{B},\) we have \(P(A)=0.4, P(B)=0.3,\) and \(P(A\) and \(B)=0.1.\) Find \(P(\operatorname{not} B).\)

Short Answer

Expert verified
The probability of event B not occurring, represented as \(P(\operatorname{not} B)\), is 0.7.

Step by step solution

01

Understand the given values

The given probabilities are \(P(A) = 0.4\), \(P(B) = 0.3\), and \(P(A \operatorname{and} B) = 0.1\). The goal is to find \(P(\operatorname{not} B)\), which is the probability of event B not happening.
02

Apply the rule of probability for not-event

According to the rules of probability, the sum of the probabilities of an event and its not-event equals 1. In other words, \(P(B) + P(\operatorname{not} B) = 1\). Thus it follows that \(P(\operatorname{not} B) = 1 - P(B)\).
03

Compute the value

Substitute the given value for \(P(B) = 0.3\) into the formula we derived in Step 2. By doing this, we find \(P(\operatorname{not} B) = 1 - P(B) = 1 - 0.3 = 0.7\). Therefore, the probability that event B does not occur is 0.7.

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Key Concepts

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

Conditional Probability
Imagine that you're trying to understand the likelihood of an event, knowing that another event has already occurred. This is where conditional probability comes in. **Conditional probability** deals with the probability of event A happening given that event B has occurred. It's noted as \( P(A | B) \). A key formula for calculating this is:
  • \( P(A | B) = \frac{P(A \cap B)}{P(B)} \)
where \( P(A \cap B) \) is the probability that both events A and B occur, and \( P(B) \) is the probability of event B occurring. Remember, this formula only works if \( P(B) > 0 \). Conditional probability is useful in fields like insurance and finance, where predictions based on existing knowledge are necessary.
Understanding how probability adjusts due to existing conditions helps in decision-making and assessing risk. It's essential to not confuse it with simply multiplying the individual probabilities of two independent events.
Complementary Events
When discussing probability, you often hear about complementary events. But what exactly are they? Complementary events are two outcomes that cover all possibilities of a given situation. If one happens, the other cannot, and vice versa. For example, if event B is passing an exam, then the complementary event, \( \operatorname{not} B \), would be failing the exam.
  • Complementary events have probabilities that add up to 1: \( P(B) + P(\operatorname{not} B) = 1 \).
This makes understanding these events straightforward because once you know one probability, you can easily find its complement by subtracting it from 1.
Whenever you're tasked with finding the probability of a situation not happening, remember this rule: compute the probability of the original event, and subtract it from 1. It’s a handy rule that simplifies calculations, especially in cases with many outcomes.
Rules of Probability
Probability is governed by fundamental rules that help in the calculation of complex probabilities. These **rules of probability** form the basis of calculating probabilities accurately.One of the primary rules is the **addition rule for non-mutually exclusive events**. It is used to determine the probability of either event A or event B occurring:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Here, \( P(A \cup B) \) is the probability that either event A or B (or both) occurs. The term \( P(A \cap B) \) is subtracted to avoid double accounting in non-mutually exclusive events where both can happen.
Another crucial rule is the **multiplication rule for independent events** which is used when two events do not affect each other's outcome:
  • \( P(A \cap B) = P(A) \times P(B) \)
These rules provide a framework for calculating probabilities efficiently and effectively. Understanding them helps interpret real-world situations and make informed predictions.

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Most popular questions from this chapter

In Exercises \(\mathrm{P} .74\) to \(\mathrm{P} .77\), fill in the \(?\) to make \(p(x)\) a probability function. If not possible, say so. $$ \begin{array}{lccc} \hline x & 1 & 2 & 3 \\ \hline p(x) & 0.5 & 0.6 & ? \\ \hline \end{array} $$

Use the fact that we have independent events \(\mathrm{A}\) and \(\mathrm{B}\) with \(P(A)=0.7\) and \(P(B)=0.6\). Find \(P(A\) or \(B)\).

State whether the two events (A and B) described are disjoint, independent, and/or complements. (It is possible that the two events fall into more than one of the three categories, or none of them.) South Africa plays Australia for the championship in the Rugby World Cup. Let \(\mathrm{A}\) be the event that Australia wins and \(\mathrm{B}\) be the event that South Africa wins. (The game cannot end in a tie.)

The word "free" is contained in \(4.75 \%\) of all messages, and \(3.57 \%\) of all messages both contain the word "free" and are marked as spam. (a) What is the probability that a message contains the word "free", given that it is spam? (b) What is the probability that a message is spam, given that it contains the word "free"?

Benford's Law Frank Benford, a physicist working in the 1930 s, discovered an interesting fact about some sets of numbers. While you might expect the first digits of numbers such as street addresses or checkbook entries to be randomly distributed (each with probability \(1 / 9\) ), Benford showed that in many cases the distribution of leading digits is not random, but rather tends to have more ones, with decreasing frequencies as the digits get larger. If a random variable \(X\) records the first digit in a street address, Benford's law says the probability function for \(X\) is $$ P(X=k)=\log _{10}(1+1 / k) $$ (a) According to Benford's law, what is the probability that a leading digit of a street address is \(1 ?\) What is the probability for \(9 ?\) (b) Using this probability function, what proportion of street addresses begin with a digit greater than \(2 ?\)

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