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Rolling dice Suppose you roll two fair, six-sided dice-one red and one green. Are the events "sum is 8 " and "green die shows a 4 " independent? Justify your answer.

Short Answer

Expert verified
The events are not independent because \(P(A \cap B) \neq P(A) \cdot P(B)\).

Step by step solution

01

Understanding the Concept of Independence

Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. Mathematically, events A and B are independent if \(P(A \cap B) = P(A) \cdot P(B)\).
02

Define the Events

Let event A be "the sum of the dice is 8" and event B be "the green die shows a 4". We need to determine \(P(A)\), \(P(B)\), and \(P(A \cap B)\).
03

Calculate \(P(A)\)

Event A, "sum is 8", occurs when the outcomes are (2,6), (3,5), (4,4), (5,3), or (6,2). There are 5 favorable outcomes out of a total of 36. Thus, \(P(A) = \frac{5}{36}\).
04

Calculate \(P(B)\)

Event B, "green die shows a 4", occurs when the green die is 4, regardless of the red die's outcome. There are 6 favorable outcomes out of 36, so \(P(B) = \frac{1}{6}\).
05

Calculate \(P(A \cap B)\)

Event "A and B" occurs when the outcome is (4,4). There is 1 favorable outcome, so \(P(A \cap B) = \frac{1}{36}\).
06

Check Independence

Check if \(P(A \cap B) = P(A) \cdot P(B)\). We have \(P(A \cap B) = \frac{1}{36}\) and \(P(A) \cdot P(B) = \frac{5}{36} \cdot \frac{1}{6} = \frac{5}{216}\). Since \(\frac{1}{36} eq \frac{5}{216}\), events A and B are not independent.

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Key Concepts

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

Independent Events
When dealing with probability, two events are considered independent if the outcome of one event does not influence the outcome of another. This means that the occurrence of one event does not change the probability of the other event happening. For example, when rolling two dice, the result of the first die does not affect the result of the second die. This is a classic case of independent events.

To determine if two events, A and B, are independent, we use the formula: \[ P(A \cap B) = P(A) \cdot P(B) \]
If this equation holds true, the events are independent. If not, they are dependent events. It is crucial to understand this concept as it lays the foundation for more complex probability problems.
Probability of Events
The probability of an event is a measure of the chance that the event will occur. Probabilities are expressed as numbers between 0 and 1, where 0 indicates the event cannot happen, and 1 signifies certainty that the event will occur. To calculate the probability of a specific event, you can use the formula:\[ P(\text{event} ) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]
In the context of dice rolling, this means counting the number of ways a certain result can occur and dividing it by the total number of possible results.
  • For example, the probability of rolling a sum of 8 with two dice is determined by counting the pairs of numbers that add up to 8 and dividing by 36, which is the total number of possible outcomes when two dice are rolled.
  • Similarly, if we want the probability of the green die showing a 4, we only consider the number of ways the green die can show 4, irrespective of the red die’s result, divided by 6 (the number of sides on a die).
Rolling Dice
Rolling dice is a classic method for introducing probability, as each die has a finite number of evenly distributed outcomes. Each roll of a die presents an independent event because every face has an equal chance of being displayed regardless of previous results.

In the scenario where you roll two six-sided dice, there are 36 combinations in total, calculated as \(6 \times 6\). Each die functions independently, meaning the result of one roll does not affect the result of another. This nature of dice throws makes them an ideal model for probability exercises, helping us explore more complex topics like the total sum of numbers or conditional probabilities. You can calculate specific events, such as getting a sum of 8, by counting combinations like (2,6), (3,5), etc.
Conditional Probability
Conditional probability deals with determining the probability of an event occurring, given that another event has already occurred. It is expressed with the notation \( P(A | B) \), meaning "the probability of A given B." This is crucial when events are not independent, as it factors in that the occurrence of event B may impact the likelihood of event A.

While the discussed example does not showcase conditional probability directly, understanding this concept helps examine relationships between dependent events. For example, if you know a green die showed a 4, this affects the calculation of probabilities related to the total dice sum, as not all combinations are possible anymore. This can offer valuable insights when decisions rely on preceding events, embodying real-world scenarios of dependent conditions.

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Most popular questions from this chapter

Free downloads? Illegal music downloading has become a big problem: \(29 \%\) of Internet users download music files, and \(67 \%\) of downloaders say they don't care if the music is copyrighted. \({ }^{17}\) What percent of Internet users download music and don't care if it's copyrighted? Write the information given in terms of probabilities, and use the general multiplication rule.

An unenlightened gambler (a) A gambler knows that red and black are equally likely to occur on each spin of a roulette wheel. He observes five consecutive reds occur and bets heavily on black at the next spin. Asked why, he explains that black is "due by the law of averages." Explain to the gambler what is wrong with this reasoning. (b) After hearing you explain why red and black are still equally likely after five reds on the roulette wheel, the gambler moves to a poker game. He is dealt five straight red cards. He remembers what you said and assumes that the next card dealt in the same hand is equally likely to be red or black. Is the gambler right or wrong, and why?

Fill 'er up! In a recent month, \(88 \%\) of automobile drivers filled their vehicles with regular gasoline, \(2 \%\) purchased midgrade gas, and \(10 \%\) bought pre- mium gas. \({ }^{18}\) Of those who bought regular gas, \(28 \%\) paid with a credit card; of customers who bought midgrade and premium gas, \(34 \%\) and \(42 \%\), respectively, paid with a credit card. Suppose we select a customer at random. (a) Draw a tree diagram to represent this situation. (b) Find the probability that the customer paid with a credit card. Show your work. (c) Given that the customer paid with a credit card, find the probability that she bought premium gas. Show your work.

Inspecting switches A shipment contains 10,000 switches. Of these, 1000 are bad. An inspector draws 2 switches at random, one after the other. (a) Draw a tree diagram that shows the sample space of this chance process. (b) Find the probability that both switches are defective.

Monty Hall problem In Parade magazine, a reader posed the following question to Marilyn vos Savant and the "Ask Marilyn" column: Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say \(\\# 1,\) and the host, who knows what's behind the doors, opens another door, say # \(3,\) which has a goat. He says to you, "Do you want to pick door # \(2 ?\) Is it to your advantage to switch your choice of doors? \(^{4}\) The game show in question was Let's Make a Deal and the host was Monty Hall. Here's the first part of Marilyn's response: "Yes; you should switch. The first door has a \(1 / 3\) chance of winning, but the second door has a \(2 / 3\) chance." Thousands of readers wrote to Marilyn to disagree with her answer. But she held her ground. (a) Use an online Let's Make a Deal applet to perform at least 50 repetitions of the simulation. Record whether you stay or switch (try to do each about half the time) and the outcome of each repetition. (b) Do you agree with Marilyn or her readers? Explain.

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