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

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

Two events are independent if the outcome of one does not affect the outcome of the other. When rolling two separate dice, each die has 6 faces numbered from 1 to 6, and the outcome of one die does not affect the outcome of the other. Therefore, the outcomes on the green and red dice are independent.
02

Calculating Probability of Specific Outcome

The probability of rolling any specific number on one die is 1/6. For outcomes (5 on green die and 3 on red die), since the outcomes are independent, multiply the probabilities:\[ P(5 \text{ on green die and } 3 \text{ on red die}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}. \]
03

Calculating Probability of another Outcome

Similarly, the probability of rolling a 3 on the green die and 5 on the red die is:\[ P(3 \text{ on green die and } 5 \text{ on red die}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}. \]
04

Calculating Probability of Either Outcome

Since these two outcomes are mutually exclusive (they cannot happen at the same time), add the probabilities of the two events:\[ P((5 \text{ on green die and } 3 \text{ on red die}) \text{ or } (3 \text{ on green die and } 5 \text{ on red die})) = \frac{1}{36} + \frac{1}{36} = \frac{2}{36} = \frac{1}{18}. \]

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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 key concept in probability that simplify computations by allowing probabilities to be multiplied. If you roll two dice, the roll of one die does not influence the roll of the other.
  • Each die has six sides, numbered from 1 to 6.
  • The result from the green die does not change the likelihood of any outcome on the red die.
  • This independence means we can calculate the probability of events involving both dice by multiplying the probability of individual events.
For example, the probability of rolling a 5 on the green die and a 3 on the red die is the product of these independent probabilities. Both dice have a 1 in 6 chance for a specific side to show up. Thus, we compute this probability as \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \). This calculation holds true for any combination of two numbers rolled on two dice as long as they are independent.
Mutually Exclusive Events
In probability, mutually exclusive events are scenarios that cannot occur at the same time. Consider two events involving dice: rolling a 5 on the green die and a 3 on the red die, and rolling a 3 on the green die and a 5 on the red die.
  • A die cannot show two different numbers at the same time.
  • For our dice scenario, either one outcome occurs or the other, but not both simultaneously.
  • This makes the events mutually exclusive.
When events are mutually exclusive, calculating the probability of either event occurring is straightforward: simply add the probabilities of each event. For the dice example, \( \frac{1}{36} \) for 5 on green and 3 on red, plus \( \frac{1}{36} \) for 3 on green and 5 on red, gives a combined probability of \( \frac{1}{18} \). This principle is crucial in solving probability problems involving multiple distinct scenarios.
Dice Probabilities
Dice probabilities provide a simple yet effective way to understand fundamental probability concepts. Each die rolled is an independent event with a uniform probability distribution.
  • There are six possible outcomes for each die roll.
  • Each outcome is equally likely, with a probability of \( \frac{1}{6} \).
  • When calculating probabilities, whether for a specific outcome or combination, we use these basic properties.
Given that the probability for one specific outcome on a single die is \( \frac{1}{6} \), calculating the probability of two specific outcomes, one for each die, is the multiplication of their probabilities due to their independence. For example, the probability for one die to show 3 and the other 5 is \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \). Utilizing these straightforward rules can tackle most basic probability problems involving dice, paving the way for more complex scenarios in mathematics.

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Most popular questions from this chapter

Probability Estimate: Wiggle Your Ears Can you wiggle your ears? Use the students in your statistics class (or a group of friends) to estimate the percentage of people who can wiggle their ears. How can your result be thought of as an estimate for the probability that a person chosen at random can wiggle his or her ears? Comment: National statistics indicate that about \(13 \%\) of Americans can wiggle their ears (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York).

Greg made up another question for a small quiz. He assigns the probabilities \(P(A)=0.6, P(B)=0.7, P(A | B)=0.1\) and asks for the probability \(P(A \text { or } B\) ). What is wrong with the probability assignments?

What is the law of large numbers? If you were using the relative frequency of an event to estimate the probability of the event, would it be better to use 100 trials or 500 trials? Explain.

Consider a family with 3 children. Assume the probability that one child is a boy is 0.5 and the probability that one child is a girl is also \(0.5,\) and that the events "boy" and "girl" are independent. (a) List the equally likely events for the gender of the 3 children, from oldest to youngest. (b) What is the probability that all 3 children are male? Notice that the complement of the event "all three children are male" is "at least one of the children is female." Use this information to compute the probability that at least one child is female.

Counting Four wires (red, green, blue, and yellow) need to be attached to a circuit board. A robotic device will attach the wires. The wires can be attached in any order, and the production manager wishes to determine which order would be fastest for the robot to use. Use the multiplication rule of counting to determine the number of possible sequences of assembly that must be tested. Hint: There are four choices for the first wire, three for the second, two for the third, and only one for the fourth.

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