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Chapter 12: Problem 23

George and Kassanthra play a game in which they roll two unbiased cubical dice. The first one who rolls a sum of 6 wins. Kassanthra rolls the dice first. a) What is the probability that Kassanthra wins on her second roll? b) What is the probability that George wins on his second roll? c) What is the probability that Kassanthra wins?

Short Answer

Expert verified
a) \( \frac{155}{1296} \), b) \( \frac{2965}{46656} \), c) \( \frac{481}{1296} \).

Step by step solution

01

Probability of Rolling a Sum of 6

First, we calculate the probability of rolling a sum of 6 with two dice. The possible outcomes that sum to 6 are: (1,5), (2,4), (3,3), (4,2), and (5,1). These are 5 successful outcomes. Since each die has 6 faces, there are a total of 36 possible outcomes when two dice are rolled (6 x 6). Therefore, the probability is \( \frac{5}{36} \).
02

Calculating Kassanthra's Probability of Winning on Second Roll

For Kassanthra to win on her second roll, she must not roll a sum of 6 on her first roll, and then roll a sum of 6 on her second roll. The probability that she does not roll a 6 on her first roll is \( 1 - \frac{5}{36} = \frac{31}{36} \). Thus, the probability she wins on her second roll is \( \frac{31}{36} \times \frac{5}{36} = \frac{155}{1296} \).
03

Calculating George's Probability of Winning on Second Roll

George can only roll if Kassanthra does not win on both her first and her second roll. So, Kassanthra must not roll a 6 on either roll. The probability of not rolling a 6 on both rolls is \( \left(\frac{31}{36}\right)^2 = \frac{961}{1296} \). For George to win on his second roll, he must lose on his first roll and win on his second, so the probability for this is \( \frac{961}{1296} \times \frac{31}{36} \times \frac{5}{36} = \frac{2965}{46656} \).
04

Calculating Kassanthra's Overall Probability of Winning

Kassanthra wins by either rolling a 6 on her first roll or by not rolling a 6 and then rolling it on subsequent rounds before George can win. Summing the probabilities of winning on the first and second rolls: the probability of winning on the first roll is \( \frac{5}{36} \), and the probability on the second roll is \( \frac{155}{1296} \). Thus, the total probability is \( \frac{5}{36} + \frac{155}{1296} = \frac{481}{1296} \).

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Key Concepts

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

Sum of Dice Outcomes
When George and Kassanthra play their game, they are interested in the sum of the numbers rolled on two dice. Each die is cubical, with numbers from 1 to 6 on its faces. This means there are a total of 36 different outcomes when both dice are rolled together, as each die has 6 possibilities. The goal is to roll a total sum of 6.

Let's discover how this can happen:
  • (1, 5)
  • (2, 4)
  • (3, 3)
  • (4, 2)
  • (5, 1)
These five pairs are the only ones that will give a sum of 6. Therefore, the probability of rolling a sum of 6 with two dice is calculated by dividing the number of successful outcomes by the total number of outcomes, which is \( \frac{5}{36} \). Understanding this helps establish the foundation for determining who can win in George and Kassanthra's game.
Rolling Cubical Dice
Rolling two unbiased cubical dice means that each face of the dice has an equal chance of showing up. This characteristic is crucial for probability in games because it ensures fairness, meaning no face is more likely to appear than another.

Here is how this works in practice:
  • Each die has 6 faces numbered from 1 to 6.
  • When rolling two dice, every roll results in a pair of numbers like (1,2) or (4,6).
  • With two dice, there are \(6 \times 6 = 36\) total possible outcomes.
Each outcome pair is equally likely, which is what unbiased implies. This principle is essential for calculating the probabilities of different sums, such as 6, which are crucial in deciding the winner in the game.
Probability Calculations
Calculating probabilities can seem complicated, but it becomes simple when broken into steps. Each step relies on understanding the likelihood of certain events happening.

Kassanthra's and George's probabilities of winning change based on their rolls:
  • To win on her second roll, Kassanthra must first fail to roll a 6, followed by succeeding on her second try. The probability of not rolling a 6 is \( \frac{31}{36} \), and winning on the second attempt combines these probabilities: \( \frac{31}{36} \times \frac{5}{36} = \frac{155}{1296} \).
  • For George to win on his second roll, Kassanthra must not win in her first two attempts. If this happens, George needs to lose on his first roll but succeed on the second. This results in: \( \frac{961}{1296} \times \frac{31}{36} \times \frac{5}{36} = \frac{2965}{46656} \).
Understanding each player's probability can help predict the game's possible outcomes. Analyzing these scenarios through careful calculations allows us to see how the game might proceed.

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Most popular questions from this chapter

In a sample space \(S\), we have the following events and some associated probabilities: $$P(E)=\frac{2}{3}, P(A | E)=\frac{3}{50} \text { and } P\left(A | E^{\prime}\right)=\frac{1}{25}$$ a) Represent the information using a tree diagram. b) Find \(P(A)\) c) Find \(P(E | A)\)

A construction company is bidding on three projects: \(B_{1}, B_{2}\) and \(B_{3}\). From previous experience they have the following probabilities of winning the bids: \(P\left(B_{1}\right)=0.22, P\left(B_{2}\right)=0.25\) and \(P\left(B_{3}\right)=0.28 .\) Winning the bids is not independent one from another. The joint probabilities are given below. $$\begin{array}{|c|c|c|c|} \hline & B_{1} & B_{2} & B_{3} \\ \hline B_{1} & & 0.11 & 0.05 \\ \hline B_{2} & 0.11 & & 0.07 \\ \hline B_{3} & 0.05 & 0.07 & \\ \hline \end{array}$$ Also, \(P\left(B_{1} \cap B_{2} \cap B_{3}\right)=0.01 .\) Find the following probabilities: a) \(P\left(B_{1} \cup B_{2}\right)\) b) \(P\left(B^{\prime}_{1} \cap B_{2}^{\prime}\right)\) c) \(P\left(B^{\prime}, \cap B_{2}^{\prime}\right) \cup B_{3}\) d) \(P\left(B^{\prime}_{1} \cap B_{2}^{\prime} \cap B_{3}\right)\) e) \(P\left(B_{2} \cap B_{3} | B_{1}\right)\) \(f) P\left(B_{2} \cup B_{3} | B_{1}\right)\)

Car insurance companies categorize drivers as high risk, medium risk and low risk. (For your information only: Teens and seniors are considered high risk!) \(20 \%\) of the drivers insured by 'First Insurance'are high risk, and \(50 \%\) are medium risk driver. The company's actuaries estimate the chance that each class of driver will have at least one accident in the coming 12 months as follows: High risk \(6 \%\) medium \(3 \%\) and low \(1 \%\) a) Find the probability that a randomly chosen driver is a high-risk driver that will have an accident in a 12 -month period. b) Find the probability that a randomly chosen driver insured by this company will have an accident in the next 12 months. c) A customer has a claim for an accident. What is the probability that he/she is a high-risk driver?

Two dice are rolled and the numbers on the top face are observed. a) List the elements of the sample space. b) Let \(x\) represent the sum of the numbers observed. Copy and complete the following table. $$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|} \hline x & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(x) & & \frac{1}{18} & & & & & & & & & \\ \hline \end{array}$$ c) What is the probability that at least one die shows a \(6 ?\) d) What is the probability that the sum is at most \(10 ?\) e) What is the probability that a die shows 4 or the sum is \(10 ?\) f) Given that the sum is \(10,\) what is the probability that one of the dice is a \(4 ?\)

A die is constructed in a way that a 1 has the chance to occur twice as often as any other number. a) Find the probability that a 5 appears. b) Find the probability an odd number will occur.
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