Problem 48 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

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

Step 1: (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.

Step 2: (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.

Step 3: (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.