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Give the multiplicative rule of probability for a. two independent events. b. any two events.

Short Answer

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a. \(P(A \cap B) = P(A) \times P(B)\); b. \(P(A \cap B) = P(A) \times P(B|A)\).

Step by step solution

01

Understanding the Multiplicative Rule for Independent Events

For two events, say \(A\) and \(B\), to be independent, the occurrence of one does not affect the probability of the occurrence of the other. The multiplicative rule for independent events states:\[P(A \cap B) = P(A) \times P(B)\]Here, \(P(A \cap B)\) represents the probability of both events \(A\) and \(B\) happen simultaneously.
02

Understanding the Multiplicative Rule for Any Two Events

For any two events \(A\) and \(B\), whether they are independent or not, the multiplicative rule involves conditional probability. The probability that events \(A\) and \(B\) both occur is given by:\[P(A \cap B) = P(A) \times P(B|A)\]This means the probability of both \(A\) and \(B\) occurring is the probability of \(A\) occurring times the probability of \(B\) occurring given that \(A\) has occurred.

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Key Concepts

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

Independent Events
When we talk about independent events, we are referring to two events that do not influence each other. This is a fundamental concept in probability.
  • For instance, if you flip a coin and roll a die, the outcome of the coin does not change the outcome of the die.
  • This lack of influence means the events are independent.
The mathematical representation of this relationship is straightforward. If events \(A\) and \(B\) are independent, the probability that both events occur is simply the product of their individual probabilities. This is expressed by the formula:\[P(A \cap B) = P(A) \times P(B)\]This equation is known as the "Multiplicative Rule for Independent Events". It's a handy rule because if you're certain that two events don't affect each other, you can easily calculate the probability of them both happening at the same time. Just multiply their individual odds!Keep in mind:
  • The key to using this rule is being sure about the independence of events.
  • If there's any interaction, however small, the events might not be truly independent.
Conditional Probability
Conditional probability comes into play when the occurrence of one event affects the likelihood of another event. In other words, we are "conditioning" the probability of one event on another.To understand conditional probability, consider event \(A\), which has already occurred, and event \(B\), whose probability we want to calculate given \(A\). This is denoted as \(P(B|A)\). The vertical bar "|" stands for "given that" or "conditioned on".For example:
  • Imagine drawing two cards from a deck without replacement.
  • If the first card is an ace, the probability of drawing a second ace is affected by the first draw.
This relationship is expressed in the multiplicative rule for any two events, say \(A\) and \(B\), by the equation:\[P(A \cap B) = P(A) \times P(B|A)\]This tells us that the probability of both events \(A\) and \(B\) happening is the product of the probability of \(A\) and the probability of \(B\) occurring given that \(A\) has already taken place. This highlights how important it is to know whether events affect each other when calculating probabilities.Remember, conditional probability helps us refine our expectations based on new information, making it crucial in many real-world applications.
Probability of Intersection
The probability of intersection is simply the probability that two events \(A\) and \(B\) occur together. This is represented as \(P(A \cap B)\). This concept is crucial in understanding scenarios where simultaneous occurrences of two events are considered.In simple terms, finding the probability of an intersection helps answer questions like:
  • What is the chance that I will pick a red card and roll a six simultaneously?
  • What is the likelihood of it raining and having a power outage on the same day?
This probability can be calculated differently based on whether the events are independent or not:- **For independent events**: Use \(P(A \cap B) = P(A) \times P(B)\). Since no event influences the other, their probabilities multiply straightforwardly.- **For dependent events**: Use \(P(A \cap B) = P(A) \times P(B|A)\). Here, we factor in the impact of one event on the probability of the other.Understanding the probability of intersection fully is key to mastering more complex probability concepts. It's like fitting puzzle pieces together to see the bigger picture of potential outcomes. This concept not only helps in theory but also in practical applications, like assessing risk or planning decisions under uncertainty.

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Most popular questions from this chapter

A salesperson living in city \(A\) wishes to visit four cities \(B, C, D,\) and \(E\) a. If the cities are all connected by airlines, how many different travel plans could be constructed to visit each city exactly once and then return home? b. Suppose all cities are connected, except that \(B\) and \(C\) are not directly connected. How many different flight plans would be available to the salesperson?

In October 1994 , a flaw was discovered in the Pentium microchip installed in personal computers. The chip produced an incorrect result when dividing two numbers. Intel, the manufacturer of the Pentium chip, initially announced that such an error would occur once in 9 billion division operations, or "once every 27,000 years," for a typical user; consequently, Intel did not immediately offer to replace the chip. Depending on the procedure, statistical software packages (e.g., SAS) may perform an extremely large number of divisions to produce required output. For heavy users of the software, 1 billion divisions over a short time frame is not unusual. Will the flawed chip be a problem for a heavy SAS user? [Note: Two months after the flaw was discovered, Intel agreed to replace all Pentium chips free of charge.]

Explain the difference between an unconditional probability and a conditional probability.

Dominant versus recessive traits. An individual's genetic makeup is determined by the genes obtained from each parent. For every genetic trait, each parent possesses a gene pair, and each parent contributes one-half of this gene pair, with equal probability, to his or her offspring, forming a new gene pair. The offspring's traits (eye color, baldness, etc.) come from this new gene pair, each gene of which possesses some characteristic. For the gene pair that determines eye color, each gene trait may be one of two types: dominant brown \((B)\) or recessive blue \((b)\). A person possessing the gene pair \(B B\) or \(B b\) has brown eyes, whereas the gene pair \(b b\) produces blue eyes. a. Suppose both parents of an individual are brown eyed, each with a gene pair of type \(B b\). What is the probability that a randomly selected child of this couple will have blue eyes? b. If one parent has brown eyes, type \(B b\), and the other has blue eyes, what is the probability that a randomly selected child of this couple will have blue eyes? c. Suppose one parent is brown eyed with a gene pair of type \(B B\). What is the probability that a child has blue eyes?

The genetic origin and properties of maize (modern-day corn) were investigated in Economic Botany. Seeds from maize ears carry either single spikelets or paired spikelets, but not both. Progeny tests on approximately 600 maize ears revealed the following information: Forty percent of all seeds carry single spikelets, while \(60 \%\) carry paired spikelets. A seed with single spikelets will produce maize ears with single spikelets \(29 \%\) of the time and paired spikelets \(71 \%\) of the time. A seed with paired spikelets will produce maize ears with single spikelets \(26 \%\) of the time and paired spikelets \(74 \%\) of the time. a. Find the probability that a randomly selected maize ear seed carries a single spikelet and produces ears with single spikelets. b. Find the probability that a randomly selected maize ear seed produces ears with paired spikelets.

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