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A 12 -sided die \(A\) has 9 green faces and 3 white faces, whereas another 12 -sided die \(B\) has 3 green faces and 9 white faces. A fair coin is tossed once. If it falls heads, a series of throws is made with die \(A\) alone; if it falls tails then only the die \(B\) is used. (a) Show that the probability that green turns up at the first throw is \(\frac{1}{2}\). (b) If green turns up at the first throw, what is the probability that die \(A\) is being used? (c) Given that green turns up at the first two throws, what is the probability that green turns up at the third throw?

Short Answer

Expert verified
(a) \(\frac{1}{2}\), (b) \(\frac{3}{4}\), (c) \(\frac{3}{4}\)

Step by step solution

01

Determine the Probability for Die Outcome

First, calculate the probability of getting a green face on the first roll for each die. For die \(A\), the probability is \(\frac{9}{12} = \frac{3}{4}\). For die \(B\), it is \(\frac{3}{12} = \frac{1}{4}\).
02

Calculate the Total Probability of Green on First Throw

Since a coin determines which die is used, with heads choosing die \(A\) and tails choosing die \(B\), calculate the combined probability of a greenface. Using the law of total probability: \(P(G) = P(G|A)P(A) + P(G|B)P(B) \). With each coin side being equally probable,\[ P(G) = \frac{3}{4} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{1}{2} = \frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \].
03

Probability of Using Die A Given Green Result

Use Bayes' theorem to find the probability that die \(A\) is used if the first throw results in green: \[ P(A|G) = \frac{P(G|A) \cdot P(A)}{P(G)} = \frac{\frac{3}{4} \cdot \frac{1}{2}}{\frac{1}{2}} = \frac{3}{4} \].
04

Probability of Green on Third Throw Given Two Greens

Given green turns up twice, calculate the probability for a third. Since Die A was likely used, and its green probability is \(\frac{3}{4}\), \[ P(G_3|G_1 \cap G_2) = P(G_3|A) = \frac{3}{4} \].

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Key Concepts

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

Understanding Conditional Probability
Conditional probability is a fundamental concept in probability theory. It's the probability of an event occurring, given that another event has already occurred. In our exercise, we're particularly interested in the outcome when a green face shows up on a first throw.
Suppose we know the coin flip resulted in using die \(A\). The conditional probability is then \(P(G|A) = \frac{9}{12} = \frac{3}{4}\). Similarly, if die \(B\) is used, \(P(G|B) = \frac{3}{12} = \frac{1}{4}\).
By knowing which die is in play, conditional probability changes how we look at outcomes. It helps us determine the likeliness of an event based on the occurrence of another. This is crucial for understanding the dynamics when working with multiple scenarios or experiments.
Exploring Bayes' Theorem
Bayes' theorem deals with updating probabilities based on new information. In our context, it helps measure how likely it is for die \(A\) to be used if green is the result.
The formula for Bayes' theorem is:
  • \(P(A|G) = \frac{P(G|A) \cdot P(A)}{P(G)}\)
Here, \(P(A|G)\) is the probability of using die \(A\) given that the throw is green. In our solution, we see that the theorem simplifies the calculation to \(\frac{3}{4}\). Updating probabilities in this way allows us to revise our understanding as events unfold. It provides us a quantitative mechanism to process new evidence or data.
Applying the Law of Total Probability
The law of total probability is a tool used to determine the overall probability of an event when accounting for several separate cases or "events". When a coin flip dictates whether we use die \(A\) or \(B\), we need to consider both options to find the probability of rolling green.
For our problem, the law is applied as follows:
  • \(P(G) = P(G|A)P(A) + P(G|B)P(B)\)
Since each side of the coin is equally probable, you find the overall probability of green on the first roll as \(\frac{1}{2}\).
This law is helpful when thinking about situations with multiple paths to a result. It's like building a bridge between distinct events and understanding the entire scenario.
Deciphering Dice Probabilities
Dice probabilities refer to the likelihood of different outcomes when rolling dice. Our exercise uses two 12-sided dice, each with distinct compositions of green and white faces.
For die \(A\), which has 9 green and 3 white faces, the probability of green is \(\frac{3}{4}\) per roll. For die \(B\), with 3 green and 9 white faces, that probability is \(\frac{1}{4}\).
When analyzing dice outcomes, we consider both individual possibilities and their combinations in different scenarios. Knowing these probabilities helps in evaluating the chances of different series of events and is vital in strategy-based games and certain statistical problems. Understanding how each die works paves the way to predict outcomes accurately, which is the essence of probability theory when applied to dice.

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Most popular questions from this chapter

A team of three students Amy, Bella, and Carol answer questions in a quiz. A question is answered by Amy, Bella, or Carol with probability \(\frac{1}{2}, \frac{1}{3}\), or \(\frac{1}{6}\), respectively. The probability of Amy, Bella, or Carol answering a question correctly is \(\frac{4}{5}\), \(\frac{3}{5}\), or \(\frac{3}{5}\), respectively. What is the probability that the team answers a question correctly? Find the probability that Carol answered the question given that the team answered incorrectly. The team starts the contest with one point and gains (loses) one point for each correct (incorrect) answer. The contest ends when the team's score reaches zero points or 10 points. Find the probability that the team will win the contest by scoring 10 points, and show that this is approximately \(\frac{4}{7}\).

You have to play Alekhine, Botvinnik, and Capablanca once each. You win each game with respective probabilities \(p_{a}, p_{b}\), and \(p_{c}\), where \(p_{a}>p_{b}>p_{c}\). You win the tournament if you win two consecutive games, otherwise you lose, but you can choose in which order to play the three games. Show that to maximize your chance of winning you should play Alekhine second.

A network forming the edges of a cube is constructed using 12 wires, each 1 metre long. An ant is placed on one corner and walks around the network, leaving a trail of scent as it does so. It never turns around in the middle of an edge, and when it reaches a corner: (i) If it has previously walked along both the other edges, it returns along the edge on which it has just come. (ii) If it has previously walked along just one of the other edges, it continues along the edge along which it has not previously walked. (iii) Otherwise, it chooses one of the other edges arbitrarily. Show that the probability that the ant passes through the corner opposite where it started after walking along just three edges is \(\frac{1}{2}\), but that it is possible that it never reaches the opposite corner. In the latter case, determine the probability of this occurring. What is the greatest distance that the ant has to walk before an outside observer (who knows the rules) will know whether the ant will ever reach the corner opposite where it started? Show that the rules may be modified to guarantee that the ant (whose only sense is smell) will be able to reach the corner opposite the corner where it started by walking not more than a certain maximum distance that should be determined. The ant can count.

Two roads join Ayton to Beaton, and two further roads join Beaton to the City. Ayton is directly connected to the City by a railway. All four roads and the railway are each independently blocked by snow with probability \(p\). I am at Ayton. (a) Find the probability that I can drive to the City. (b) Find the probability that I can travel to the City. (c) Given that I can travel to the City, what is the probability that the railway is blocked?

Suppose that parents are equally likely to have (in total) one, two, or three offspring. A girl is selected at random; what is the probability that the family includes no older girl? (Assume that children are independent and equally likely to be male or female.)

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