/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Four balls are selected at rando... [FREE SOLUTION] | 91Ó°ÊÓ

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Four balls are selected at random without replacement from an urn containing three white balls and five blue balls. Find the probability of the given event. Two or three of the balls are white.

Short Answer

Expert verified
The probability that two or three of the selected balls will be white is \(\frac{1}{2}\).

Step by step solution

01

Case 1 - Exactly two selected balls are white

In this case, if two balls are selected balls are white, then the other two selected balls must be blue. We would need to find the number of ways to choose 2 white balls out of 3 available and 2 blue balls out of 5 available.
02

Calculate combinations for Case 1

In order to calculate the number of combinations, we will use the combinations formula: \[C(n, k) = \frac{n!}{k!(n-k)!},\] where \(n\) represents the total number of items and \(k\) represents the number of items to be selected. For 2 white balls out of 3: \[C(3, 2) = \frac{3!}{2!(3-2)!} = 3\] For 2 blue balls out of 5: \[C(5, 2) = \frac{5!}{2!(5-2)!} = 10\]
03

Calculate total combinations for Case 1

Now, we need to multiply these two combinations to find the total combinations for Case 1 (2 white and 2 blue balls): \[3 \times 10 = 30\]
04

Case 2 - Exactly three selected balls are white

In this case, if three balls are selected balls are white, then the other selected ball must be blue. We would need to find the number of ways to choose 3 white balls out of 3 available and 1 blue ball out of 5 available.
05

Calculate combinations for Case 2

Using the same combinations formula: For 3 white balls out of 3: \[C(3, 3) = \frac{3!}{3!(3-3)!} = 1\] For 1 blue ball out of 5: \[C(5, 1) = \frac{5!}{1!(5-1)!} = 5\]
06

Calculate total combinations for Case 2

Now, we need to multiply these two combinations to find the total combinations for Case 2 (3 white and 1 blue ball): \[1 \times 5 = 5\]
07

Calculate total combinations for both cases

We need to add up the total combinations for both cases: \[30 (Case 1) + 5 (Case 2) = 35\]
08

Find total possible combinations

To find the total possible combinations of picking 4 balls out of 8 (3 white and 5 blue), we use the combinations formula: \[C(8, 4) = \frac{8!}{4!(8-4)!} = 70\]
09

Calculate the probability

Now, we need to divide the total combinations of both cases by the total possible combinations: \[\frac{35}{70} = \frac{1}{2}\] The probability that two or three of the selected balls will be white is \(\frac{1}{2}\).

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arranging, and finding patterns in sets. It's especially useful when you need to understand the number of ways things can happen. In the problem at hand, we're using combinatorics to find out how many ways we can select balls from an urn. For example, to find out how many ways we can select 2 white balls from 3, or 2 blue balls from 5, we use combinations. The formula for combinations, given by \(C(n, k) = \frac{n!}{k!(n-k)!}\), is designed to provide the number of *un-ordered* outcomes possible. The symbol \(n!\) denotes the factorial of \(n\), which is the product of all positive integers up to \(n\). This principle is crucial in determining the number of ways we can select items where order doesn't matter. Understanding combinatorics helps us quickly solve problems that initially might seem complex.
Mathematics Problem Solving
Mathematics problem solving often involves breaking a complex problem into smaller, manageable parts. This allows us to solve each part step by step, making the entire problem more approachable. In this urn example, we face a problem of determining probabilities across different conditions. First, we establish distinct cases: exactly 2 white balls or exactly 3. This separation is key to simplifying and understanding the problem, letting us apply specific mathematical methods to each case. Then, we use our previously discussed combinations to calculate possible outcomes for each case. The final step is solving a straightforward division of successful outcomes against total possible outcomes, culminating in the desired probability. This structured approach can be applied to countless mathematical scenarios, demonstrating the versatility and power of organized problem-solving.
Finite Mathematics
Finite mathematics is focused on understanding and solving problems where the set of possible outcomes is limited or countable, like in our urn example with a specific number of balls. This domain includes topics such as combinatorics and probability and applies to problems that involve a limited range of possibilities. The key here is recognizing that we're working within a finite framework of only 8 balls total, meaning we can enumerate all possible selections. By clearly defining the scope from the beginning, we maintain a control mechanism over our calculations. Finite mathematics is not only precise but also incredibly practical, especially in scenarios like computing probable events or outcomes within a controlled environment. Understanding this concept allows for more logical and efficient problem-solving, minimizing errors and simplifying seemingly complex situations.

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Most popular questions from this chapter

Asurvey involving 400 likely Democratic voters and 300 likely Republican voters asked the question: Do you support or oppose legislation that would require registration of all handguns? The following results were obtained: $$\begin{array}{lcc} \hline \text { Answer } & \text { Democrats, \% } & \text { Republicans, \% } \\\ \hline \text { Support } & 77 & 59 \\ \hline \text { Oppose } & 14 & 31 \\ \hline \text { Don't know/refused } & 9 & 10 \\ \hline \end{array}$$ If a randomly chosen respondent in the survey answered "oppose," what is the probability that he or she is a likely Democratic voter?

In a survey conducted in the fall 2006, 800 homeowners were asked about their expectations regarding the value of their home in the next few years; the results of the survey are summarized below: $$\begin{array}{lc} \hline \text { Expectations } & \text { Homeowners } \\ \hline \text { Decrease } & 48 \\ \hline \text { Stay the same } & 152 \\ \hline \text { Increase less than } 5 \% & 232 \\ \hline \text { Increase 5-10\% } & 240 \\ \hline \text { Increase more than 10\% } & 128 \\ \hline \end{array}$$ If a homeowner in the survey is chosen at random, what is the probability that he or she expected his or her home to a. Stay the same or decrease in value in the next few years? b. Increase \(5 \%\) or more in value in the next few years?

The probabilitics that the three patients who are scheduled to receive kidney transplants at General Hospital will suffer rejection are \(\frac{1}{2}, \frac{1}{3}\) and \(\frac{1}{10}\). Assuming that the cvents (kidney rejection) are indcpendent, find the probability that a. At least one paticnt will suffer rejection. b. Exactly two patients will suffer rejection.

In the game of blackjack, a 2 -card hand consisting of an ace and a face card or a 10 is called a blackjack. a. If a player is dealt 2 cards from a standard deck of 52 well-shuffled cards, what is the probability that the player will receive a blackjack? b. If a player is dealt 2 cards from 2 well-shuffled standard decks, what is the probability that the player will receive a blackjack?

According to the Centers for Disease Control and Prevention, the percentage of adults \(25 \mathrm{yr}\) and older who smoke, by educational level, is as follows: $$\begin{array}{lcccccc} \hline & & & \text { High } & {\text { Under- }} \\ \text { Educational } & \text { No } & \text { GED } & \text { school } & \text { Some } & \text { graduate } & \text { Graduate } \\ \text { Level } & \text { diploma } & \text { diploma } & \text { graduate } & \text { college } & \text { level } & \text { degree } \\ \hline \text { Respondents, \% } & 26 & 43 & 25 & 23 & 10.7 & 7 \\ \hline \end{array}$$ In a group of 140 people, there were 8 with no diploma, 14 with GED diplomas, 40 high school graduates, 24 with some college, 42 with an undergraduate degree, and 12 with a graduate degree. (Assume that these categories are mutually exclusive.) If a person selected at random from this group was a smoker, what is the probability that he or she is a person with a graduate degree?

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