A fair coin is tossed until two tails occur successively. Find the expected number of the tosses required. Hint: Let X = B 1 if the first toss results in tails 0 if the first toss results in heads, and condition on X.
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Textbook Solutions for Fundamentals of Probability, with Stochastic Processes
Question
In a box, Lynn has b batteries of which d are dead. She tests them randomly and one by one. Every time that a good battery is drawn, she will return it to the box; every time that a dead battery is drawn, she will replace it by a good one. (a) Determine the expected value of the number of good batteries in the box after n of them are checked. (b) Determine the probability that on the nth draw Lynn draws a good battery. Hint: Let Xn be the number of good batteries in the box after n of them are checked. Show that E(Xn | Xn1) = 1 + * 1 1 b , Xn1. Then, by computing the expected value of this random variable, find a recursive relation between E(Xn) and E(Xn1). Use this relation and induction to prove that E(Xn) = b d * 1 1 b ,n . Note that n should approach to get E(Xn) = b. For part (b), let En be the event that on the nth draw she gets a good battery. By conditioning on Xn1 prove that P (En) = E(Xn1)/b.
Solution
The first step in solving 10.4 problem number 3 trying to solve the problem we have to refer to the textbook question: In a box, Lynn has b batteries of which d are dead. She tests them randomly and one by one. Every time that a good battery is drawn, she will return it to the box; every time that a dead battery is drawn, she will replace it by a good one. (a) Determine the expected value of the number of good batteries in the box after n of them are checked. (b) Determine the probability that on the nth draw Lynn draws a good battery. Hint: Let Xn be the number of good batteries in the box after n of them are checked. Show that E(Xn | Xn1) = 1 + * 1 1 b , Xn1. Then, by computing the expected value of this random variable, find a recursive relation between E(Xn) and E(Xn1). Use this relation and induction to prove that E(Xn) = b d * 1 1 b ,n . Note that n should approach to get E(Xn) = b. For part (b), let En be the event that on the nth draw she gets a good battery. By conditioning on Xn1 prove that P (En) = E(Xn1)/b.
From the textbook chapter Conditioning on Random Variables you will find a few key concepts needed to solve this.
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