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Tyler and Gebriella are among seven contestants from which four semifinalists are to be selected at random. Find the probability that Tyler but not Gèbriella is selected.

Short Answer

Expert verified
The probability that Tyler is selected among the semifinalists and not Gabriella is \(\frac{2}{7}\).

Step by step solution

01

Determine the total number of ways the semifinalists can be selected

From the total of 7 contestants, we need to select 4 semifinalists. We can use the combination formula to find the total number of ways to choose 4 semifinalists: \[ C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} \] In this case, n = 7 (total number of contestants), and k = 4 (total number of semifinalists): \[ C(7, 4) = \binom{7}{4} = \frac{7!}{4!3!} = 35 \] There are 35 different ways to choose the 4 semifinalists from the 7 contestants.
02

Determine the number of ways to choose the semifinalists with Tyler and without Gabriella

Since Tyler is one of the semifinalists, we are now left with choosing 3 more semifinalists from the remaining 5 contestants (considering Gabriella is not chosen). Using the combination formula again, we have: \[ C(5, 3) = \binom{5}{3} = \frac{5!}{3!2!} = 10 \] There are 10 different ways to choose the 3 semifinalists from the remaining 5 contestants, including Tyler and excluding Gabriella.
03

Calculate the probability of selecting Tyler but not Gabriella

Now, we will calculate the probability by dividing the number of ways to choose the semifinalists with Tyler and without Gabriella by the total number of ways to choose the semifinalists: \[ Probability = \frac{number\ of\ favorable\ outcomes}{total\ number\ of\ outcomes} = \frac{10}{35} = \frac{2}{7} \] The probability that only Tyler is selected among the semifinalists and not Gabriella is \(\frac{2}{7}\).

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Key Concepts

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

Combination Formula
In combinatorics, the combination formula is a fundamental tool used to determine the number of ways to choose a subset of items from a larger set without considering the order of selection. This concept is crucial when faced with problems asking us to find the number of possible groups or selections where arrangement doesn't matter.

The formula for combinations is given by \[C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \(C(n, k)\) denotes the number of combinations (or ways) to choose \(k\) items from a set of \(n\) items, \(n!\) is the factorial of \(n\), \(k!\) is the factorial of \(k\), and \((n-k)!\) is the factorial of \(n-k\).

For example, if you have a group of 7 unique items and want to know how many ways you can select 4 of them, you apply the combination formula. Using factorials, which we'll discuss in the next section, you carry out the necessary calculations to arrive at the total possible combinations.
Factorial
The factorial, represented by the exclamation mark (!), plays a key role in various mathematical concepts, including combinatorics and probability. The factorial of a non-negative integer \(n\) is the product of all positive integers less than or equal to \(n\).

Formally, the factorial is defined as:\[n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1\]For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Importantly, by definition, the factorial of 0 is 1, or \(0! = 1\).

Factorials become particularly large very quickly with increasing \(n\), and they are essential when solving problems involving permutations and combinations because they provide the total count of possible arrangements or selections.
Event Probability
Event probability is a measure of how likely it is for a particular event to occur within a set of all possible outcomes. Probabilities are expressed as a number between 0 and 1, where 0 means the event cannot happen and 1 means the event is certain to happen.

The probability of an event \(A\) occurring is calculated using the formula:\[Probability(A) = \frac{\text{Number of ways event A can occur}}{\text{Total number of possible outcomes}}\].

For instance, when flipping a fair coin, the probability of getting heads is \(\frac{1}{2}\) because there are two possible outcomes (heads or tails), and only one of these is the desired event (heads). Understanding how to calculate event probability is indispensable for making predictions based on mathematical models.
Combinatorial Probability
Combinatorial probability involves calculating the likelihood of an event by considering the various combinations of outcomes that can occur in a random experiment. It combines the principles of combinations and event probability, and it is particularly useful when dealing with problems where you are asked to find the probability of a specific selection from a group.

To calculate combinatorial probability, you first determine the number of favorable outcomes (using combinations) and then divide it by the total number of possible outcomes. For example, if you want to find the probability that a certain individual is selected in a random draw from a group, as in the Tyler and Gabriella scenario, you would use the combination formula to determine both the favorable outcomes and the total outcomes, then apply the event probability formula to get the final probability.

By mastering combinatorial probability, students can analyze complex problems in games, sports, and various real-world scenarios to predict outcomes and make informed decisions based on likelihoods.

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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." Find all pairs of events that are not mutually exclusive among the events \(P, E, I, V, N\), and \(D .\)

You are given a transition matrix \(P\) and initial distribution vector \(v\). Find (a) the two-step transition matrix and (b) the distribution vectors after one, two, and three steps. $$ P=\left[\begin{array}{lll} .5 & .5 & 0 \\ 0 & 1 & 0 \\ 0 & .5 & .5 \end{array}\right], v=[1\\} $$

Complete the following probability distribution table, and then calculate the stated probabilities. $$ \begin{array}{|r|c|c|c|c|c|} \hline \text { Outcome } & \mathrm{a} & \mathrm{b} & \mathrm{c} & \mathrm{d} & \mathrm{e} \\ \hline \text { Probability } & .1 & .05 & .6 & .05 & \\ \hline \end{array} $$ a. \(P(\\{a, c, e)\\}\) b. \(P(E \cup F),\) where \(E=\\{\mathrm{a}, \mathrm{c}, \mathrm{e}\\}\) and \(F=\\{\mathrm{b}, \mathrm{c}, \mathrm{e}\\}\) c. \(P\left(E^{\prime}\right),\) where \(E\) is as in part (b) d. \(P(E \cap F)\), where \(E\) and \(F\) are as in part (b)

Concern the following chart, which shows the way in which a dog moves its facial muscles when torn between the drives of fight and flight. \({ }^{4}\) The "fight" drive increases from left to right; the "fight" drive increases from top to bottom. (Notice that an increase in the "fight" drive causes its upper lip to lift, while an increase in the "flight" drive draws its ears downward.) \(\nabla\) Let \(E\) be the event that the dog's flight drive is the strongest, let \(F\) be the event that the dog's flight drive is weakest, let \(G\) be the event that the dog's fight drive is the strongest, and let \(H\) be the event that the dog's fight drive is weakest. Describe the following events in terms of \(E, F, G\), and \(H\) using the symbols \(\cap, \cup\), and \(^{\prime} .\) a. The dog's flight drive is weakest and its fight drive is not weakest. b. The dog's flight drive is not strongest or its fight drive is weakest. c. Either the dog's flight drive or its fight drive fails to be strongest.

Based on the following table, which shows U.S. employment figures for 2007, broken down by educational attainment. \(^{49}\) All numbers are in millions, and represent civilians aged 25 years and over. Those classed as "not in labor force " were not employed nor actively seeking employment. Round all answers to two decimal places. Your friend claims that a person not in the labor force is more likely to have less than a high school diploma than an employed person. Respond to this claim by citing actual probabilities.

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