A box contains three marbles: one red, one green, and one blue. Consider an experiment that consists of taking one marble from the box then replacing it in the box and drawing a second marble from the box. What is the sample space? If, at all times, each marble in the box is equally likely to be selected, what is the probability of each point in the sample space?

Short Answer

Expert verified
The sample space for the experiment is {RR, RG, RB, GR, GG, GB, BR, BG, BB}. The probability of each point in the sample space is \(\frac{1}{9}\).

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Identify the possible outcomes

: To determine the sample space, we will list all possible combinations of drawing two marbles from the box, taking into account the replacement after the first draw: 1. Red, Red (RR) 2. Red, Green (RG) 3. Red, Blue (RB) 4. Green, Red (GR) 5. Green, Green (GG) 6. Green, Blue (GB) 7. Blue, Red (BR) 8. Blue, Green (BG) 9. Blue, Blue (BB) These 9 combinations represent the sample space for this experiment: Sample Space = {RR, RG, RB, GR, GG, GB, BR, BG, BB}
02

Calculate the probability of each point in the sample space

: Since there are 3 marbles in the box and at all times each marble has an equal chance to be selected, the probability of selecting any particular marble in each draw is \(\frac{1}{3}\). As the marbles are replaced after each draw, the draws are independent events. Remember that the probability of independent events occurring together is the product of their individual probabilities. Now, we will calculate the probability of each point in the sample space: 1. P(RR) = P(Red on 1st Draw) * P(Red on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 2. P(RG) = P(Red on 1st Draw) * P(Green on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 3. P(RB) = P(Red on 1st Draw) * P(Blue on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 4. P(GR) = P(Green on 1st Draw) * P(Red on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 5. P(GG) = P(Green on 1st Draw) * P(Green on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 6. P(GB) = P(Green on 1st Draw) * P(Blue on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 7. P(BR) = P(Blue on 1st Draw) * P(Red on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 8. P(BG) = P(Blue on 1st Draw) * P(Green on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) 9. P(BB) = P(Blue on 1st Draw) * P(Blue on 2nd Draw) = \(\frac{1}{3} * \frac{1}{3}\) = \(\frac{1}{9}\) The probabilities of each point in the sample space are: P(RR) = P(RG) = P(RB) = P(GR) = P(GG) = P(GB) = P(BR) = P(BG) = P(BB) = \(\frac{1}{9}\)

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free