/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 A fair die is successively rolle... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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) \(E[X | Y=1]\) (c) \(E[X | Y=5]\)

Short Answer

Expert verified
(a) \(E[X] = 6\) (b) \(E[X|Y=1] = 6\) (c) \(E[X|Y=5] = 11\)

Step by step solution

01

(a) Find E[X]

To find the expected number of rolls needed to obtain a 6 (E[X]), we can consider the geometric distribution. Since the die is fair, the probability of rolling a 6, denoted as p, is equal to 1/6. Therefore, we can find the expected value using the formula E[X] = 1/p. \(E[X] = \frac{1}{\frac{1}{6}} = 6\) The expected number of rolls needed to obtain a 6 is 6.
02

(b) Find E[X | Y = 1]

We'll next find the expected number of rolls needed to obtain a 6 given that a 5 has already been rolled on the first roll (E[X | Y = 1]). Since we already know that the first roll was a 5, there are 5 possible outcomes remaining (1, 2, 3, 4, or 6). The probability of rolling a 6 after the first roll is now 1/5. We can use the geometric distribution formula again, but now conditioned on Y=1. \(E[X|Y=1] = 1 + \frac{1}{\frac{1}{5}} = 1 + 5\) The expected number of rolls needed to obtain a 6 given that a 5 was rolled on the first roll is 6.
03

(c) Find E[X | Y = 5]

Finally, we'll find the expected number of rolls needed to obtain a 6 given that a 5 was rolled on the fifth roll (E[X | Y = 5]). Since the rolls are independent, having a 5 on the fifth roll doesn't change the probability of getting a 6 on any roll. The only new information we have from Y=5 is that we didn't get a 6 in the first 4 rolls, but this doesn't change the probability of rolling a 6 on any of the following rolls. Hence, the probability of getting a 6 after 5 rolls is still 1/6 and we can use the geometric distribution formula again, but now conditioned on Y=5. \(E[X|Y=5] = 5 + \frac{1}{\frac{1}{6}} = 5 + 6\) The expected number of rolls needed to obtain a 6 given that a 5 was rolled on the fifth roll is 11.

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Most popular questions from this chapter

Consider a gambler who, at each gamble, either wins or loses her bet with respective probabilities \(p\) and \(1-p .\) A popular gambling system known as the Kelley strategy is to always bet the fraction \(2 p-1\) of your current fortune when \(p>\frac{1}{2}\) Compute the expected fortune after \(n\) gambles of a gambler who starts with \(x\) units and employs the Kelley strategy.

Each of \(m+2\) players pays 1 unit to a kitty in order to play the following game: A fair coin is to be flipped successively \(n\) times, where \(n\) is an odd number, and the successive outcomes are noted. Before the \(n\) llips, each player writes down a prediction of the outcomes. For instance, if \(n=3\) then a player might write down \((H, H, T),\) which means that he or she predicts that the first flip will land on heads, the second on heads, and the third on tails. After the coins are flipped, the players count their total number of correct predictions. Thus, if the actual outcomes are all heads, then the player who wrote \((H, H, T)\) would have 2 correct predictions. The total kitty of \(m+2\) is then evenly split up among those players having the largest number of correct predictions. since each of the coin flips is equally likely to land on either heads or tails, \(m\) of the players have decided to make their predictions in a totally random fashion. Specifically, they will each flip one of their own fair coins \(n\) times and then use the result as their prediction. However, the final 2 of the players have formed a syndicate and will use the following strategy: One of them will make predictions in the same random fashion as the other \(m\) players, but the other one will then predict exactly the opposite of the first. That is, when the randomizing member of the syndicate predicts an \(H,\) the other member predicts a \(T .\) For instance, if the randomizing member of the syndicate predicts \((H, H, T),\) then the other one predicts \((T,\) \(T, H)\) (a) Argue that exactly one of the syndicate members will have more than \(n / 2\) correct predictions. (Remember, \(n\) is odd.) (b) Let \(X\) denote the number of the \(m\) nonsyndicate players that have more than \(n / 2\) correct predictions. What is the distribution of \(X ?\) (c) With \(X\) as defined in part (b), argue that \(E[\text { payoff to the syndicate }]=(m+2)\) $$ \times E\left[\frac{1}{X+1}\right] $$(d) Use part (c) of Problem 59 to conclude that \(\begin{aligned} E[\text { payoff to the syndicate }]=& \frac{2(m+2)}{m+1} \\\ & \times\left[1-\left(\frac{1}{2}\right)^{m+1}\right] \end{aligned}\) and explicitly compute this number when \(m=\) \(1,2,\) and \(3 .\) Because it can be shown that $$ \frac{2(m+2)}{m+1}\left[1-\left(\frac{1}{2}\right)^{m+1}\right]>2 $$ it follows that the syndicate's strategy always gives it a positive expected profit.

If \(X_{1}, X_{2}, \ldots, X_{n}\) are independent and identically distributed random variables having uniform distributions over \((0,1),\) find (a) \(E\left[\max \left(X_{1}, \ldots, X_{n}\right)\right]\) (b) \(E\left[\min \left(X_{1}, \ldots, X_{n}\right)\right]\)

How many times would you expect to roll a fair die before all 6 sides appeared at least once?

A fair die is rolled 10 times. Calculate the expected sum of the 10 rolls.

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