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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?

Short Answer

Expert verified
(a) \( \frac{1}{6} \); (b) \( \frac{1}{18} \).

Step by step solution

01

Understanding the Problem

Dick throws a die and the result of the first throw determines the number of additional throws he will make. Dick throws the die a total of \( j + (j-1) = 2j - 1 \) times. We are asked to find the probability that the sum of all these throws is exactly 3.
02

Considering One Throw Only

First, we consider the scenario where Dick only throws the die once. In this case, the result of the throw, \( j \), must itself be 3, as it is the only score contributing to the total sum. Therefore, Dick must roll a die result of \( 3 \) on his very first throw.
03

Probability for One Throw

The probability of rolling a 3 on a fair six-sided die is \( \frac{1}{6} \).
04

Considering Two Throws

Now, consider the scenario where Dick throws the die twice. This occurs when \( j = 2 \), because then he throws it \( j - 1 = 1 \) more time. Thus, the results \( a \) and \( b \) of the two throws must sum to 3: \( a + b = 3 \). Possible pairs \((a, b)\) that satisfy this are: \((2, 1)\), \((1, 2)\).
05

Probability for Two Throws

First, Dick must roll a 2 on the first throw, which occurs with probability \( \frac{1}{6} \). Then, for each sequence of pairs: \( (2, 1) \) or \( (1, 2) \), the total additional probability that matches the second throw is \( \frac{1}{6} \), making the combined probability for either sequence \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \). Adding the probability of each valid sequence together, the probability for rolling \((2, 1)\) or \((1, 2)\) in either order is \( 2 \times \frac{1}{36} = \frac{2}{36} \).
06

Conclusion

So the probabilities are: (a) Probability that Dick throws the die only once is \( \frac{1}{6} \), and (b) Probability that Dick throws the die twice is \( \frac{2}{36} = \frac{1}{18} \).

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Key Concepts

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

Understanding Conditional Probability in Dice Throws
Conditional probability is a key concept when throwing dice under specific conditions like in this problem with Dick. In simple terms, conditional probability is the likelihood of an event happening given another event has already occurred.
In Dick's case, if he rolls a specific number on his first die throw, this number determines how many more times he will roll the dice. Therefore, calculating the probability depends on the initial outcome of the first die throw and what follows.
  • First Roll Impact: Dick’s first roll directly influences subsequent actions. Initially, he rolls the die, and based on that number, he rolls again "(j – 1)" times.
  • Condition Example: If his first roll is a 2, then he will roll it one more time because "2 – 1 = 1." This shows directly how the initial outcome applies a condition to subsequent events.
When the sum of the scores equals 3, understanding this dependency between rolls is crucial. It allows us to use conditional probability to determine the chances that apply based on the initial roll results.
Calculating the Probability of Rolling Specific Totals
Probability calculation involves determining the likelihood of specific outcomes. In the dice scenario, the goal is to find probabilities for specific scenarios where the sums of the dice match given conditions.
For Dick's game:
  • Single Roll Probability: If Dick throws the dice only once, the only way to achieve a sum of 3 is if the first roll results in 3. Therefore, the basic probability calculation for this scenario is straightforward: probability of rolling a 3 = \( \frac{1}{6} \).
  • Double Roll Probability: If he rolls twice, the possibilities increase. Initially, the first roll could be 2, and the subsequent throws would need to sum to 1. The combination of outcomes (2,1) or (1,2) matters here.
For calculating the probabilities in these scenarios, we take into account the probability of each rolling event happening in sequence and how likely each needed number is to appear when the die is thrown this number of times.
Exploring Discrete Random Variables in Dice Rolls
A discrete random variable is a variable that can take on a finite number of values. In the case of dice, each face of the die represents a possible value, making it a perfect example of a discrete random variable.
  • Each die throw can land on a value from 1 to 6, each equally likely, forming the discrete sample space.
  • For Dick's problem, once we establish how many times the die is rolled, each roll becomes a discrete event tied to specific probabilities.
When assessing probabilities like the ones we calculated earlier, understanding that dice outcomes are discrete allows easier configuration of possible results.
In practical terms, Dick's potential results for each throw are finite and countable, leading to clear probability calculations for every scenario of multiple throws, examining potential sequences that lead to the desired total of 3. This highlights the entire structure of the problem within the framework of discrete random variables.

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Most popular questions from this chapter

An urn contains four dice, one red, one green, and two blue. (a) One is selected at random; what is the probability that it is blue? (b) The first is not replaced, and a second die is removed. What is the chance that it is: (i) blue? or (ii) red? (c) The two dice are thrown. What is the probability that they show the same numbers and are the same colour? (d) Now the two remaining in the urn are tossed. What is the probability that they show the same number and are the same colour, given that the first two did not show the same number and colour?

Box \(A\) contains three red balls and two white balls; box \(B\) contains two red balls and two white balls. A fair die is thrown. If the upper face of the die shows 1 or 2 , a ball is drawn at random from box \(A\) and put in box \(B\) and then a ball is drawn at random from box \(B\). If the upper face of the die shows \(3,4,5\) or 6 , a ball is drawn at random from box \(B\) and put in box \(A\), and then a ball is drawn at random from box \(A\). What are the probabilities (a) That the second ball drawn is white? (b) That both balls drawn are red? (c) That the upper face of the red die showed 3, given that one ball drawn is white and the other red?

Let \(A\) and \(B\) be independent events. Show that $$ \max \left\\{\mathbf{P}\left((A \cup B)^{c}\right), \mathbf{P}(A \cap B), \mathbf{P}(A \Delta B)\right\\} \geq \frac{4}{9} $$

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 $$ s_{n}=\frac{1}{2}\left(1+(p-q)^{n}\right) ; \quad n \geq 1 $$

\(A\) 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?

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