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Chapter 4: Problem 9

You roll two fair dice, a green one and a red one. (a) Are the outcomes on the dice independent? (b) Find \(P(5\) on green die and 3 on red die). (c) Find \(P(3\) on green die and 5 on red die). (d) Find \(P((5\) on green die and 3 on red die) or \((3\) on green die and 5 on red die \()\).

Short Answer

Expert verified
(a) Yes, the outcomes are independent. (b) \(\frac{1}{36}\). (c) \(\frac{1}{36}\). (d) \(\frac{1}{18}\).

Step by step solution

01

Understanding Independence

For two events to be independent, the occurrence of one event does not affect the occurrence of the other. In the case of rolling two dice, whether the first die shows a 5 or not has no impact on what the second die will show. Therefore, the outcomes on the two dice are independent.
02

Calculating Individual Probabilities

When rolling a fair die, each side has a probability of \(\frac{1}{6}\) of landing face up. Therefore, the probability of rolling a 5 on the green die is \(\frac{1}{6}\), and the probability of rolling a 3 on the red die is also \(\frac{1}{6}\).
03

Calculating Joint Probability for 5 on Green and 3 on Red

Since the events are independent, the joint probability of rolling a 5 on the green die and a 3 on the red die is the product of their individual probabilities: \[ P(5 \text{ on green and } 3 \text{ on red}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}. \]
04

Calculating Joint Probability for 3 on Green and 5 on Red

Using the same logic as the step above, the probability of rolling a 3 on the green die and a 5 on the red die is also: \[ P(3 \text{ on green and } 5 \text{ on red}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}. \]
05

Calculating Probability of Either Combination

To find the probability of one event occurring or the other (not both simultaneously), we add the probabilities of the two independent events. Therefore, the probability that either the green die shows 5 and the red shows 3, or the green shows 3 and the red shows 5 is: \[ P((5 \text{ on green and } 3 \text{ on red}) \text{ or } (3 \text{ on green and } 5 \text{ on red})) = \frac{1}{36} + \frac{1}{36} = \frac{2}{36} = \frac{1}{18}. \]

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Key Concepts

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

Independent Events
Independent events are a crucial concept in probability that simplifies calculations and helps us analyze different situations easily. When two events are independent, it means that the occurrence of one event has absolutely no effect on the probability of the other event occurring. This is exactly what happens when you roll two dice.
  • One die landing on a specific number doesn't change the outcome of the other die.
  • This lack of influence makes dice rolls classic examples of independent events.
Understanding event independence is fundamental because it helps in determining joint probabilities, which we'll look into next.
Joint Probability
Joint probability is a term used to describe the probability of two or more events occurring simultaneously. When dealing with independent events, calculating joint probability is straightforward.
  • For independent events, the joint probability is simply the product of the probabilities of each individual event.
  • For example, the probability of rolling a 5 on one die and a 3 on another die would be calculated by multiplying the probability of each event happening independently.
This product rule simplifies joint probability calculations, especially in cases involving a series of independent actions.
Rolling Dice
Rolling dice is a fantastic real-world context for understanding probability concepts, especially for beginners. Each die has six faces, numbered from 1 to 6, and every side has an equal chance of landing face up when the die is rolled.
  • Each outcome is equally likely, which means any number on a standard die has a probability of \(\frac{1}{6}\).
  • This probability remains the same with each roll, no matter how the previous rolls may have turned out.
This feature of fairness in dice makes them perfect tools for exploring the concept of independent events and understanding likelihood in probability.
Probability Calculation
Probability calculation often involves straightforward arithmetic when dealing with events like dice rolling. Knowing how to calculate probabilities of combined events is essential.
  • To find the probability of getting a 5 on the green die and a 3 on the red die, you multiply their individual probabilities: \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).
  • Likewise, the probability of getting a 3 on the green die and a 5 on the red die is also \(\frac{1}{36}\).
When looking for the probability of either event occurring, you add the probabilities of both the independent events: \(\frac{2}{36} = \frac{1}{18}\). This simple approach to probability calculations ensures clarity and accuracy in understanding outcomes.

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Most popular questions from this chapter

Arches National Park is located in southern Utah. The park is famous for its beautiful desert landscape and its many natural sandstone arches. Park Ranger Edward McCarrick started an inventory (not yet complete) of natural arches within the park that have an opening of at least 3 feet. The following table is based on information taken from the book Canyon Country Arches and Bridges, by F. A. Barnes. The height of the arch opening is rounded to the nearest foot. $$\begin{array}{l|lllll} \hline \text { Height of arch, feet } & 3-9 & 10-29 & 30-49 & 50-74 & 75 \text { and higher } \\ \hline \begin{array}{l} \text { Number of arches } \\ \text { in park } \end{array} & 111 & 96 & 30 & 33 & 18 \\ \hline \end{array}$$ For an arch chosen at random in Arches National Park, use the preceding information to estimate the probability that the height of the arch opening is (a) 3 to 9 feet tall (d) 10 to 74 feet tall (b) 30 feet or taller (e) 75 feet or taller (c) 3 to 49 feet tall

(a) If you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes? Are the outcomes equally likely? (b) Assign probabilities to the outcomes of the sample space of part (a). Do the probabilities add up to 1 ? Should they add up to \(1 ?\) Explain. (c) What is the probability of getting a number less than 5 on a single throw? (d) What is the probability of getting 5 or 6 on a single throw?

M\&M plain candies come in various colors. According to the M\&M/Mars Department of Consumer Affairs (link to the Mars company web site from the Brase/Brase statistics site at college.hmco.com/pic/braseUS9e), the distribution of colors for plain M\&M candies is \(\begin{array}{l|ccccccc} \hline \text { Color } & \text { Purple } & \text { Yellow } & \text { Red } & \text { Orange } & \text { Green } & \text { Blue } & \text { Brown } \\ \hline \text { Percentage } & 20 \% & 20 \% & 20 \% & 10 \% & 10 \% & 10 \% & 10 \% \\ \hline \end{array}\) Suppose you have a large bag of plain M\&M candies and you choose one candy at random. Find (a) \(P(\) green candy or blue candy). Are these outcomes mutually exclusive? Why? (b) \(P(\) yellow candy or red candy). Are these outcomes mutually exclusive? Why? (c) \(P(\) not purple candy \()\)

The state medical school has discovered a new test for tuberculosis. (If the test indicates a person has tuberculosis, the test is positive.) Experimentation has shown that the probability of a positive test is \(0.82\), given that a person has tuberculosis. The probability is \(0.09\) that the test registers positive, given that the person does not have tuberculosis. Assume that in the general population, the probability that a person has tuberculosis is \(0.04\). What is the probability that a person chosen at random will (a) have tuberculosis and have a positive test? (b) not have tuberculosis? (c) not have tuberculosis and have a positive test?

If two events are mutually exclusive, can they occur concurrently? Explain.
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