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A sample space contains 10 simple events: \(E_{1}\), \(E_{2}, \ldots, E_{10} .\) If \(P\left(E_{1}\right)=3 P\left(E_{2}\right)=.45\) and the remaining simple events are equiprobable, find the probabilities of these remaining simple events.

Short Answer

Expert verified
Answer: The probability of each of the remaining simple events, \(E_3\) to \(E_{10}\), is 0.05.

Step by step solution

01

Find the probabilities of events \(E_1\) and \(E_2\)

We are given the probability of event \(E_1\) as \(P(E_1) = 3P(E_2) = 0.45\). We can find the probability of event \(E_2\) by dividing both sides of the equation by 3: $$ P(E_2) = \frac{P(E_1)}{3} = \frac{0.45}{3} $$ Calculating the value, we get: $$ P(E_2) = 0.15 $$
02

Calculate the sum of probabilities for events \(E_1\) and \(E_2\)

Now that we have the probabilities for events \(E_1\) and \(E_2\), let's find their sum: $$ P(E_1) + P(E_2) = 0.45 + 0.15 = 0.6 $$
03

Determine the probability for the remaining events

Since the sum of probabilities for all simple events in a sample space must equal 1, we can find the sum of probabilities for the remaining simple events as follows: $$ 1 - (P(E_1) + P(E_2)) = 1 - 0.6 = 0.4 $$ Since events \(E_3, E_4, \ldots, E_{10}\) are equiprobable, we can find their individual probabilities by dividing the sum of their probabilities by the number of events (8 in this case): $$ P(E_3) = P(E_4) = \ldots = P(E_{10}) = \frac{0.4}{8} $$ Calculating the value, we get: $$ P(E_3) = P(E_4) = \ldots = P(E_{10}) = 0.05 $$ Therefore, the probabilities of the remaining simple events are all equal to 0.05.

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Key Concepts

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

Sample Space
In probability theory, understanding the concept of a sample space is crucial. The sample space is essentially the collection of all possible outcomes of a probabilistic experiment. Think of it as a "universe" of sorts, where each event happens as a unique occurrence. Each outcome is called a simple event, which is a single member within the sample space.
For example, if you were to flip a coin, the sample space would include two simple events: heads and tails. In the original exercise, our sample space includes 10 simple events, expressed as \(E_1, E_2, \ldots, E_{10}\).
To fully understand a scenario, one must consider all events within the sample space since this allows for complete and accurate probability calculations. Knowing the entirety of the sample space ensures that every possibility has been accounted for, making the probability calculations reliable and valid.
Simple Events
Simple events are the fundamental building blocks of probability calculations. They represent individual outcomes that cannot be broken down into simpler components. Each simple event within a sample space is mutually exclusive, meaning that the occurrence of one simple event means none of the others can occur simultaneously.
In the exercise, we see that \(E_1\) and \(E_2\) have specific probabilities, which have been calculated based on given conditions. Simple events have distinct probabilities that are often initially unknown, necessitating comprehensive calculation or observation. This ensures their probabilities reflect true likelihoods when analyzed within a sample space context.
When a simple event's probability is known, it can help in finding unknown probabilities of other events, as demonstrated in the step-by-step solution. Knowing the attributes and probabilities of simple events within a sample space is vital for understanding a broader probabilistic scenario.
Equiprobable Events
Equiprobable events are those events within a sample space that have the same probability of occurring. This notion simplifies many probability calculations by allowing us to assume equal likelihoods across multiple events.
In the provided exercise, the events \(E_3, E_4, \ldots, E_{10}\) are equiprobable. This implies that each of these events has the same probability, derived from the remaining probability once individual probabilities for \(E_1\) and \(E_2\) have been calculated and subtracted from 1.
Seeing equiprobable events in action helps in understanding how individual event probabilities relate to the totality of probabilities within a sample space. When events are equiprobable, it indicates a symmetric distribution of probability, often simplifying the analysis. This condition is particularly useful when dealing with complex probabilistic models, as it reduces variability and complexity in calculations.

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Most popular questions from this chapter

Two cold tablets are unintentionally placed in a box containing two aspirin tablets. The four tablets are identical in appearance. One tablet is selected at random from the box and is swallowed by the first patient. A tablet is then selected at random from the three remaining tablets and is swallowed by the second patient. Define the following events as specific collections of simple events: a. The sample space \(S\) b. The event \(A\) that the first patient obtained a cold tablet c. The event \(B\) that exactly one of the two patients obtained a cold tablet d. The event \(C\) that neither patient obtained a cold tablet

An experiment can result in one of five equally likely simple events, \(E_{1}, E_{2}, \ldots, E_{5} .\) Events \(A, B,\) and \(C\) are defined as follows: \(A: E_{1}, E_{3}\) $$P(A)=.4$$ \(B: E_{1}, E_{2}, E_{4}, E_{5} \quad P(B)=.8\) \(C: E_{3}, E_{4}\) $$P(C)=.4$$ Find the probabilities associated with the following events by listing the simple events in each. a. \(A^{c}\) b. \(A \cap B\) c. \(B \cap C\) d. \(A \cup B\) e. \(B \mid C\) f. \(A \mid B\) g. \(A \cup B \cup C\) h. \((A \cap B)^{c}\)

A piece of electronic equipment contains six computer chips, two of which are defective. Three chips are selected at random, removed from the piece of equipment, and inspected. Let \(x\) equal the number of defectives observed, where \(x=0,1,\) or 2 . Find the probability distribution for \(x\) Express the results graphically as a probability histogram.

Past experience has shown that, on the average, only 1 in 10 wells drilled hits oil. Let \(x\) be the number of drillings until the first success (oil is struck). Assume that the drillings represent independent events. a. Find \(p(1), p(2),\) and \(p(3)\). b. Give a formula for \(p(x)\). c. Graph \(p(x)\).

A research physician compared the effectiveness of two blood pressure drugs \(A\) and \(B\) by administering the two drugs to each of four pairs of identical twins. Drug \(A\) was given to one member of a pair; drug \(B\) to the other. If, in fact, there is no difference in the effects of the drugs, what is the probability that the drop in the blood pressure reading for drug \(A\) exceeds the corresponding drop in the reading for drug \(B\) for all four pairs of twins? Suppose drug \(B\) created a greater drop in blood pressure than drug \(A\) for each of the four pairs of twins. Do you think this provides sufficient evidence to indicate that drug \(B\) is more effective in lowering blood pressure than drug \(A\) ?

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