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Chapter 7: Problem 44

A fair die is successively rolled. Let \(X\) and \(Y\) denote, respectively, the number of rolls necessary to obtain a 6 and a 5. Find (a) \(E[X]\) (b) \(E[X \mid Y=1]\); (c) \(E[X \mid Y=5]\).

Short Answer

Expert verified
The expected number of rolls needed to obtain a 6 are: (a) \(E[X] = 6\) (b) \(E[X | Y=1] = 7\) (c) \(E[X | Y=5] = 6\)

Step by step solution

01

Calculate E[X]

We are required to find the expected number of rolls to get a 6. This is a geometric distribution problem, where X follows a geometric distribution with success probability \( p=\frac{1}{6} \) Using the formula for the expected value of a geometric distribution: \( E[X] = \frac{1}{p} \) Substitute the success probability value of \( \frac{1}{6} \): \(E[X] = \frac{1}{\frac{1}{6}}=6 \)
02

Calculate E[X | Y=1]

To find E[X | Y=1], we should note that since 5 appeared on the first roll, the first roll for obtaining a 6 starts from the second roll. Let Z be X-1 so that Z denotes the number of trials needed to obtain a 6 after the first roll. As a result, Z will also follow a geometric distribution with the same success probability undefined as \( p=\frac{1}{6} \). Now, we need to find the expected value of Z and add 1 to account for the first roll: \( E[Z] = \frac{1}{p} = 6 \) \( E[X | Y=1] = E[Z] + 1 = 6 + 1 = 7 \)
03

Calculate E[X | Y=5]

To find E[X | Y=5], we should note that the rolls until obtaining the first 5 are irrelevant to the rolls needed for obtaining a 6. Therefore, the probability of getting a 6 is still the same (as if we just started rolling the die again). Thus, the expected number of rolls needed to obtain a 6 after 5 rolls will also be the same as E[X]: \( E[X | Y=5] = E[X] = 6 \)
04

Summarize Results

We have found the expected number of rolls needed to obtain a 6: (a) E[X] = 6 (b) E[X | Y=1] = 7 (c) E[X | Y=5] = 6

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Key Concepts

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

Expected Value
The expected value, often denoted as \(E[X]\), is a fundamental concept in probability theory. It represents the average or mean value that you would anticipate if you could repeat an experiment infinitely many times. In simpler terms, it's the long-run average outcome of a random variable.
For a geometric distribution, which models the number of trials until the first success, the expected value tells us the average number of trials needed for that success to occur.
  • In the case of a fair die with a probability of rolling a 6 as \( \frac{1}{6} \), the expected number of rolls to get a 6 is calculated using the formula \(E[X] = \frac{1}{p}\), where \(p\) is the probability of success.
  • Substituting \(p = \frac{1}{6}\) into the formula gives us \(E[X] = 6\). So on average, it takes 6 rolls to get a 6 when rolling a fair die.
Conditional Expectation
Conditional expectation \( E[X | Y] \) extends the idea of expected value by accounting for additional information or conditions.
In our context, it is the expected number of rolls needed to get a 6, given some condition about the rolls, such as already having rolled a 5.
  • For \( E[X | Y=1] \), since a 5 is rolled on the first try, the subsequent attempts can be seen as starting from a new first roll to achieve a 6. The expected rolls to get a 6 after the first roll of 5 would be \(E[Z] = 6\), and adding the initial roll, it becomes 7.
  • For \( E[X | Y=5] \), the rolls before getting a 5 are irrelevant. After these rolls, the probability remains unchanged, just like fresh rolls. Therefore, \(E[X|Y=5]=6\).
Conditional expectation helps us adjust our predictions based on new data or conditions.
Probability Theory
Probability theory provides a framework for modeling random events and describing uncertainty. It's the mathematical foundation for all statistics and many theories in science and engineering.
In probability, we are interested in the likelihood of various outcomes鈥攈ow probable they are to happen. The geometric distribution is one example, used when measuring the number of independent trials until a first success (like rolling a die until we get a certain number).
  • In a fair die scenario, each roll is an independent event with an equal chance of landing on any side. Therefore, the probability \(p\) for rolling a certain number (like a 6 or 5) remains constant at \(\frac{1}{6}\) for every roll.
  • This same theory justifies why, even with some prior rolls or outcomes (like obtaining a 5), the underlying probabilities of future rolls, and consequently the expected values, do not change unless explicitly altered by the condition.
Probability theory ties together these concepts by allowing us to make informed predictions about future outcomes based on known probabilities.

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Most popular questions from this chapter

Consider \(n\) independent trials, the ith of which results in a success with probability \(P_{i}\) (a) Compute the expected number of successes in the \(n\) trials - call it \(\mu\). (b) For fixed value of \(\mu_{1}\) what choice of \(P_{1}, \ldots, P_{n}\) maximizes the variance of the number of successes? (c) What choice minimizes the variance?

If \(Y=a X+b\), where \(a\) and \(b\) are constants, express the moment generating function of \(Y\) in terms of the moment generating function of \(X\).

A deck of \(n\) cards, numbered 1 through \(n\), is thoroughly shuffled so that all possible \(n !\) orderings can be assumed to be equally likely. Suppose you are to make \(n\) guesses sequentially, where the \(i\) th one is a guess of the card in position \(i\). Let \(N\) denote the number of correct guesses. (a) If you are not given any information about your earlier guesses show that, for any strategy, \(E[N]=1\). (b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that under this strategy $$ \begin{aligned} E[N] &=\frac{1}{n}+\frac{1}{n-1}+\cdots+1 \\ & \approx \int_{1}^{n} \frac{1}{x} d x=\log n \end{aligned} $$ (c) Suppose that you are told after each guess whether you are right or wrong. In this case it can be shown that the strategy that maximizes \(E[N]\) is one which keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy show that $$ \begin{aligned} E[N] &=1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !} \\ &=e-1 \end{aligned} $$

The number of accidents that a person has in a given year is a Poisson random variable with mean \(\lambda\). However, suppose that the value of \(\lambda\) changes from person to person, being equal to 2 for 60 percent of the population and 3 for the other 40 percent. If a person is chosen at random, what is the probability that he will have (a) 0 accidents and (b) exactly 3 accidents in a year? What is the conditional probability that he will have 3 accidents in a given year, given that he had no accidents the preceding year?

Let \(X\) be a normal random variable with parameters \(\mu=0\) and \(\sigma^{2}=1\) and let \(I\), independent of \(X\), be such that \(P\\{I=1\\}=\frac{1}{2}=P[I=0\\}\). Now define \(Y\) by $$ Y=\left\\{\begin{aligned} X & \text { if } I=1 \\ -X & \text { if } I=0 \end{aligned}\right. $$ In words, \(Y\) is equally likely to equal either \(X\) or \(-X\). (a) Are \(X\) and \(Y\) independent? (b) Are \(I\) and \(Y\) independent? (c) Show that \(Y\) is normal with mean 0 and variance \(1 .\) (d) Show that \(\operatorname{Cov}(X, Y)=0\).
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