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Three fair dice labelled \(A, B\), and \(C\) are rolled on to a sheet of paper. If a pair show the same number a straight line is drawn joining them. Show that the event that the line \(A B\) is drawn is independent of the event that \(B C\) is drawn. What is the probability that a complete triangle is drawn? (The dice are not colinear.)

Short Answer

Expert verified
Events are independent; Probability of a complete triangle is \( \frac{1}{36} \).

Step by step solution

01

Understand Event Definitions

Label the events as follows: Let event \( E_1 \) represent the event 'Line \( AB \) is drawn' which implies dice \( A \) and \( B \) show the same number. Similarly, let event \( E_2 \) be the event 'Line \( BC \) is drawn', meaning dice \( B \) and \( C \) show the same number. Finally, let event \( E_3 \) denote the event 'Line \( CA \) is drawn', meaning dice \( A \) and \( C \) show the same number.
02

Calculate Probability of Events

Since the dice are fair and have 6 sides, the probability that two dice show the same number is \( \frac{1}{6} \) for each pair (i.e., \( P(E_1) = \frac{1}{6}, P(E_2) = \frac{1}{6}, \) and \( P(E_3) = \frac{1}{6} \)).
03

Determine Joint Probability of Two Events

We need to find the joint probability \( P(E_1 \cap E_2) \). The outcomes where both \( E_1 \) and \( E_2 \) occur is when all three dice show the same number, which can happen in one favorable outcome per number on the dice, 6 such possible outcomes (one for each face). So, \( P(E_1 \cap E_2) = \frac{1}{36} \).
04

Check Independence

For \( E_1 \) and \( E_2 \) to be independent, we must show \( P(E_1 \cap E_2) = P(E_1)P(E_2) \). With \( P(E_1)P(E_2) = \left( \frac{1}{6} \right) \left( \frac{1}{6} \right) = \frac{1}{36} \), which equals \( P(E_1 \cap E_2) = \frac{1}{36} \). Therefore, \( E_1 \) and \( E_2 \) are independent.
05

Probability of Complete Triangle

A complete triangle is formed when all three lines are drawn, i.e., all three pairs of dice show the same number. Thus, we must have all three dice showing the same number, which has a probability of \( P(E_1 \cap E_2 \cap E_3) = \frac{1}{36} \) as there are just 6 ways (one for each face) for all three to match simultaneously.

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Key Concepts

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

Independent Events
In probability theory, events are considered independent if the occurrence of one event does not affect the occurrence of another.
For example, consider the scenario where three dice, labeled as A, B, and C, are rolled. Let's define two events:
  • Event \(E_1\): A line is drawn between dice A and B (i.e., dice A and B show the same number)
  • Event \(E_2\): A line is drawn between dice B and C (i.e., dice B and C show the same number)
These events are independent if the probability that both events happen simultaneously equals the product of their probabilities. In simpler terms, knowing whether or not A and B show the same number does not provide any information about whether B and C will show the same number.
In this exercise, the joint probability \(P(E_1 \cap E_2)\) was calculated as \(\frac{1}{36}\), and the independent probabilities \(P(E_1) \text{ and } P(E_2)\) were \(\frac{1}{6}\) each. Since their product also gives \(\frac{1}{36}\), it confirms that these events are indeed independent.
Joint Probability
Joint probability refers to the probability of multiple events happening at the same time. It's the foundation for understanding how different events interact in probability theory.
In this scenario, we were interested in the joint probability of events \(E_1\) and \(E_2\) occurring at the same time. This means we want the probability that both pairs of dice A-B and B-C show the same number simultaneously.
To find \(P(E_1 \cap E_2)\), note that all three dice (A, B, and C) need to show the same number for both lines to be drawn. There are only 6 favorable outcomes—one for each face of the die—out of a total of 216 combinations since each die has 6 faces \((6 \times 6 \times 6 = 216)\). Thus, the joint probability is \(\frac{6}{216} = \frac{1}{36}\).
Understanding joint probabilities helps in assessing the relationship between different events and predicting the likelihood of simultaneous occurrences.
Fair Dice
A fair die is a six-sided cube where each side is equally likely to land face-up when rolled. This property of uniformity means that each face has a \( \frac{1}{6} \) probability of appearing.
Fairness in dice is an essential concept in probability because it ensures that each outcome has an equal chance, which allows us to calculate probabilities with precision and confidence. In experiments with fair dice, such as rolling multiple dice, the outcomes are random, and the calculations assume that there is no bias in the dice.
  • This uniformity is critical when determining probabilities involving multiple dice.
  • In this exercise, the assumption of fairness ensures our probability calculations, such as finding matching dice faces, are accurate.
Understanding fair dice is foundational to many problems in probability theory, where this assumption helps simplify and clarify complex probabilistic scenarios.
Combinatorial Probability
Combinatorial probability involves counting techniques, which help us determine the probability of complex events by analyzing possible combinations of outcomes.
In this exercise, combinatorial reasoning helps us discern the number of ways dice can match to form lines (and ultimately a triangle).
  • The total number of possible outcomes when rolling three dice is calculated using the principle of multiplication \((6 \times 6 \times 6 = 216)\).
  • To calculate the probability of forming a complete triangle, we focus on the simple scenario where all dice show the same number, amounting to just one matching outcome per face.
There are only 6 such outcomes (one for each die face), and thus the probability is \( \frac{6}{216} = \frac{1}{36} \).
Combinatorial methods are powerful in simplifying and solving problems where counting different potential outcomes is necessary, especially in the study of probability.

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Most popular questions from this chapter

Weather Days can be sunny or cloudy. The weather tomorrow is the same as the weather today with probability \(p\), or it is different with probability \(q\), where \(p+q=1\). If it is sunny today, show that the probability \(s_{n}\) that it will be sunny \(n\) days from today satisfies $$ s_{n}=(p-q) s_{n-1}+q ; \quad n \geq 1, $$ where \(s_{0}=1\). Deduce that $$ s_{n}=\frac{1}{2}\left(1+(p-q)^{n}\right) ; \quad n \geq 1 $$

Dick throws a die once. If the upper face shows \(j\), he then throws it a further \(j-1\) times and adds all \(j\) scores shown. If this sum is 3 , what is the probability that he only threw the die (a) Once altogether? (b) Twice altogether?

\(A\) and \(B\) play a sequence of games. in each of which \(A\) has a probability \(p\) of winning and \(B\) has a probability \(q(=1-p)\) of winning. The sequence is won by the first player to achieve a lead of two games. By considering what may happen in the first two games, or otherwise, show that the probability that \(A\) wins the sequence is \(p^{2} /(1-2 p q)\). If the rules are changed so that the sequence is won by the player who first wins two consecutive games, show that the probability that \(A\) wins the sequence becomes \(p^{2}(1+q) /(1-p q)\). Which set of rules gives that weaker player the better chance of winning the sequence?

Suppose that for events \(S, A\), and \(B\), $$ \begin{aligned} \mathbf{P}(S \mid A) & \geq \mathbf{P}(S) \\ \mathbf{P}(A \mid S \cap B) & \geq \mathbf{P}(A \mid S) \\ \mathbf{P}\left(A \mid S^{c}\right) & \geq \mathbf{P}\left(A \mid S^{c} \cap B\right) \end{aligned} $$ (a) Show that, except in trivial cases, \(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid B)\). (b) Show that \(\mathbf{P}(S \mid A) \geq \mathbf{P}(A)\). (c) Show that if \((*)\) is replaced by \(\mathbf{P}(S \mid B) \geq \mathbf{P}(S)\), then \(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid A)\).

After marking the papers of a certain student, the examiners are unable to decide whether he really understands the subject or is just bluffing. They reckon that the probability that he is a bluffer is \(p, 0b\), then as \(n\) increases the probability that the student really understands tends to 1 . How many questions must the student get right to convince the examiners that it is more likely that he understands the subject than that he is bluffing?

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