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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

The probability of a flush A poker player holds a flush when all 5 cards in the hand belong to the same suit. We will find the probability of a flush when 5 cards are dealt. Remember that a deck contains 52 cards, 13 of each suit, and that when the deck is well shuffled, each card dealt is equally likely to be any of those that remain in the deck. (a) We will concentrate on spades. What is the probability that the first card dealt is a spade? What is the conditional probability that the second card is a spade given that the first is a spade? (b) Continue to count the remaining cards to find the conditional probabilities of a spade on the third, the fourth, and the fifth card given in each case that all previous cards are spades. (c) The probability of being dealt 5 spades is the product of the five probabilities you have found. Why? What is this probability? (d) The probability of being dealt 5 hearts or 5 diamonds or 5 clubs is the same as the probability of being dealt 5 spades. What is the probability of being dealt a flush?

Texas hold 'em In the popular Texas hold 'em variety of poker, players make their best five-card poker hand by combining the two cards they are dealt with three of five cards available to all players. You read in a book on poker that if you hold a pair (two cards of the same rank) in your hand, the probability of getting four of a kind is \(88 / 1000\). (a) Explain what this probability means. (b) If you play 1000 such hands, will you get four of a kind in exactly 88 of them? Explain.

Foreign-language study Choose a student in grades 9 to 12 at random and ask if he or she is studying a language other than English. Here is the distribution of results: $$\begin{array}{lccccc}\hline \text { Language: } & \text { Spanish } & \text { French } & \text { German } & \text { All others } & \text { None } \\\\\text { Probability: } & 0.26 & 0.09 & 0.03 & 0.03 & 0.59 \\\\\hline\end{array}$$ (a) What's the probability that the student is studying a language other than English? (b) What is the conditional probability that a student is studying Spanish given that he or she is studying some language other than English?

Role-playing games Computer games in which the players take the roles of characters are very popular. They go back to earlier tabletop games such as Dungeons \(\&\) Dragons. These games use many different types of dice. A four- sided die has faces with \(1,2,3,\) and 4 spots. (a) List the sample space for rolling the die twice (spots showing on first and second rolls). (b) What is the assignment of probabilities to outcomes in this sample space? Assume that the die is perfectly balanced.

Blood types All human blood can be typed as one of \(\mathrm{O}, \mathrm{A}, \mathrm{B},\) or \(\mathrm{AB},\) but the distribution of the types varies a bit with race. Here is the distribution of the blood type of a randomly chosen black American: $$\begin{array}{lcccc}\hline \text { Blood type: } & 0 & \mathrm{~A} & \mathrm{~B} & \mathrm{AB} \\\\\text { Probability: } & 0.49 & 0.27 & 0.20 & ? \\\\\hline\end{array}$$ (a) What is the probability of type \(A B\) blood? Why? (b) What is the probability that the person chosen does not have type \(A B\) blood? (c) Maria has type \(\mathrm{B}\) blood. She can safely receive blood transfusions from people with blood types \(\mathrm{O}\) and \(\mathrm{B}\). What is the probability that a randomly chosen black American can donate blood to Maria?

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