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A box contains 6 red, 6 yellow, and 6 green marbles. Construct a sample space for the experiment of randomly drawing out, with replacement, three marbles in succession and noting the color each time.

Short Answer

Expert verified
Sample space has 27 outcomes: combinations of R, Y, G for three draws.

Step by step solution

01

Understand the Problem

We need to find all possible outcomes when drawing three marbles, in sequence, from a box containing 6 red, 6 yellow, and 6 green marbles with replacement each time. Each draw can result in one of three colors.
02

Define the Sample Space for One Draw

With replacement, each draw can result in Red (R), Yellow (Y), or Green (G). For one draw, the sample space is \( S_1 = \{ R, Y, G \} \).
03

Extend the Sample Space for Three Draws

Since we draw with replacement, each draw is independent of the previous one. For three draws, each can independently result in R, Y, or G. To find the complete sample space, consider all combinations of R, Y, and G in sets of three.
04

Enumerate the Combinations

List all possible combinations of three colors where each draw follows one of the options R, Y, or G. Combinations include sequences like (R, R, R), (R, R, Y), (R, R, G) up to (G, G, G).
05

Calculate the Total Number of Outcomes

Since each draw can be one of 3 colors and we draw 3 times, there are \( 3^3 = 27 \) possible outcomes. Each of these outcomes is a combination of sequences chosen from (R, Y, G).
06

Construct the Full Sample Space

Write down all 27 sequences. This includes outcomes such as (R, Y, G), (G, G, G), and others until all 27 are listed. The sample space \( S \) for three draws is all sequences where each draw can be either R, Y, or G.

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Key Concepts

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

Probability
Probability is fundamentally the measure of how likely an event is to occur. In the context of our marble exercise, we use probability to determine the likelihood of drawing a specific sequence of marble colors. Each individual draw from the box is an event that can have three possible outcomes (Red, Yellow, Green).
When discussing probability, it's crucial to understand that we define it as the ratio of the number of favorable outcomes to the total number of possible outcomes. In the case of drawing three marbles, as each draw is independent and occurs with replacement, the probability of any specific sequence (like drawing R, Y, G) remains consistent.
  • Each draw has a probability of 1/3 for each color.
  • The probability of any specific three-marble sequence is calculated by multiplying the probabilities of each individual draw: \[(1/3) \times (1/3) \times (1/3) = 1/27\]
  • Therefore, every sequence out of the 27 sequences in the sample space has a probability of 1/27.
Understanding these principles is key to solving problems involving random events.
Combinatorics
Combinatorics is the branch of mathematics focused on counting, arranging, and combining objects. In our exercise, combinatorics comes into play when determining all possible outcomes of drawing marbles.
To find the complete sample space for drawing three marbles with replacement, we use the fundamental principle of counting. Since you have 3 independent draws and each draw offers 3 color options, you calculate the possible outcomes as:
  • First Draw: 3 choices (Red, Yellow, Green)
  • Second Draw: 3 choices (Red, Yellow, Green)
  • Third Draw: 3 choices (Red, Yellow, Green)
Mathematically, the number of sequences is expressed as:\[3 \times 3 \times 3 = 3^3 = 27\]This means there are 27 possible combinations of color sequences, reflecting the complete sample space. Combinatorics provides us the framework to work through such enumeration of possibilities efficiently, ensuring we capture every possible outcome.
Independent Events
Independent events are events where the outcome of one event does not affect the outcome of another. In the marble drawing exercise, the concept of independent events is crucial because each draw occurs with replacement.
When drawing the marbles from the box:
  • After each draw, the marble is replaced, ensuring the conditions remain constant for the next draw.
  • This replacement keeps each draw independent of the others, meaning the outcome of one does not influence the others.
  • For example, drawing a red marble first does not change the probability of drawing another red marble in the second draw.
This independence allows for a straightforward calculation of probabilities, as discussed earlier. It also simplifies our application of combinatorics since each draw resets the scenario, making each draw a distinct and separate event.

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Most popular questions from this chapter

A manufacturer examines its records over the last year on a component part received from outside suppliers. The breakdown on source (supplier \(A,\) supplier \(B\) ) and quality \((H:\) high \(, U:\) usable, \(D:\) defective \()\) is shown in the two-way contingency table. $$ \begin{array}{|c|c|c|c|} \hline & H & U & D \\ \hline A & 0.6937 & 0.0049 & 0.0014 \\ \hline B & 0.2982 & 0.0009 & 0.0009 \\ \hline \end{array} $$ The record of a part is selected at random. Find the probability of each of the following events. a. The part was defective. b. The part was either of high quality or was at least usable, in two ways: (i) by adding numbers in the table, and (ii) using the answer to (a) and the Probability Rule for Complements. c. The part was defective and came from supplier \(B\). d. The part was defective or came from supplier \(B\), in two ways: by finding the cells in the table that correspond to this event and adding their probabilities, and (ii) using the Additive Rule of Probability.

A sample space is \(S=\\{a, b, c, d, e\\} .\) Identify two events as \(U=\\{a, b, d\\}\) and \(V=\\{b, c, d\\}\). Suppose \(P(a)\) and \(P(b)\) are each 0.2 and \(P(c)\) and \(P(d)\) are each \(0.1 .\) a. Determine what \(P(e)\) must be. b. Find \(P(U)\). c. Find \(P(V)\)

The sample space of equally likely outcomes for the experiment of rolling two fair dice is \(\begin{array}{llllll}11 & 12 & 13 & 14 & 15 & 16\end{array}\) \(\begin{array}{llllll}21 & 32 & 23 & 34 & 25 & 36\end{array}\) \(\begin{array}{llllll}31 & 32 & 32 & 34 & 35 & 36\end{array}\) \(\begin{array}{llllll}41 & 42 & 43 & 44 & 45 & 46\end{array}\) \(\begin{array}{llllll}51 & 52 & 53 & 54 & 55 & 56\end{array}\) \(\begin{array}{llll}61 & 62 & 63 & 64 & 65 & 66\end{array}\) Identify the events \(N:\) the sum is at least nine, \(T\) : at least one of the dice is a two, and \(F:\) at least one of the dice is a five. a. Find \(P(N)\). b. Find \(P(N \mid P)\). c. Find \(P(N \mid T)\). d. Determine from the previous answers whether or not the events \(N\) and \(F\) are independent; whether or not \(N\) and \(T\) are.

An accountant has observed that \(5 \%\) of all copies of a particular two-part form have an error in Part I, and \(2 \%\) have an error in Part II. If the errors occur independently, find the probability that a randomly selected form will be error-free.

A basketball player makes \(60 \%\) of the free throws that he attempts, except that if he has just tried and missed a free throw then his chances of making a second one go down to only \(30 \%\). Suppose he has just been awarded two free throws. a. Find the probability that he makes both. b. Find the probability that he makes at least one. (A tree diagram could help.)

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