6. We have three light bulbs with lifetimes \( T_1, T_2, T_3 \) distributed according to Exponential(\(\lambda_1\)), Exponential(\(\lambda_2\)), Exponential(\(\lambda_3\)). In other words, for example bulb #1 will break at a random time \( T_1 \), where the distribution of this time \( T_1 \) is Exponential(\(\lambda_1\)). The three bulbs break independently of each other. The three light bulbs are arranged in series, one after the other, along a circuit—this means that as soon as one or more light bulbs fail, the circuit will break. Let \( T \) be the lifetime of the circuit—that is, the time until the circuit breaks. --- **(a)** What is the CDF of \( T \), the lifetime of the circuit? **(b)** Next, suppose that we only check on the circuit once every second (assume the times \( T_1, T_2, T_3, T \) are measured in seconds). Let \( S \) be the first time we check the circuit and see that it’s broken. For example, if the circuit breaks after 3.55 seconds, we will only observe this when 4 seconds have passed, and so \( S = 4 \). Calculate the PMF of \( S \). **(c)** Finally, suppose that instead of checking on the circuit every second, we instead do the following: after each second, we randomly decide whether to check on the circuit or not. With probability \( p \) we check, and with probability \( 1 - p \) we do not check. This decision is made independently at each time. Now let \( N \) be the number of times we check and see the circuit working. For example, if the circuit breaks at time 3.55, and our choices were to check at time 1 second, not to check at times 2 or 3 or 4, and to check at time 5, then \( N = 1 \), since the circuit was broken the 2nd time we checked. What is the PMF of \( N \)? (Hint: start by finding the joint PMF of \( N \) and \( S \). It's fine if your answer is in summation form.)
6. We have three light bulbs with lifetimes \( T_1, T_2, T_3 \) distributed according to Exponential(\(\lambda_1\)), Exponential(\(\lambda_2\)), Exponential(\(\lambda_3\)). In other words, for example bulb #1 will break at a random time \( T_1 \), where the distribution of this time \( T_1 \) is Exponential(\(\lambda_1\)). The three bulbs break independently of each other. The three light bulbs are arranged in series, one after the other, along a circuit—this means that as soon as one or more light bulbs fail, the circuit will break. Let \( T \) be the lifetime of the circuit—that is, the time until the circuit breaks. --- **(a)** What is the CDF of \( T \), the lifetime of the circuit? **(b)** Next, suppose that we only check on the circuit once every second (assume the times \( T_1, T_2, T_3, T \) are measured in seconds). Let \( S \) be the first time we check the circuit and see that it’s broken. For example, if the circuit breaks after 3.55 seconds, we will only observe this when 4 seconds have passed, and so \( S = 4 \). Calculate the PMF of \( S \). **(c)** Finally, suppose that instead of checking on the circuit every second, we instead do the following: after each second, we randomly decide whether to check on the circuit or not. With probability \( p \) we check, and with probability \( 1 - p \) we do not check. This decision is made independently at each time. Now let \( N \) be the number of times we check and see the circuit working. For example, if the circuit breaks at time 3.55, and our choices were to check at time 1 second, not to check at times 2 or 3 or 4, and to check at time 5, then \( N = 1 \), since the circuit was broken the 2nd time we checked. What is the PMF of \( N \)? (Hint: start by finding the joint PMF of \( N \) and \( S \). It's fine if your answer is in summation form.)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Transcribed Image Text:6. We have three light bulbs with lifetimes \( T_1, T_2, T_3 \) distributed according to Exponential(\(\lambda_1\)), Exponential(\(\lambda_2\)), Exponential(\(\lambda_3\)). In other words, for example bulb #1 will break at a random time \( T_1 \), where the distribution of this time \( T_1 \) is Exponential(\(\lambda_1\)). The three bulbs break independently of each other.
The three light bulbs are arranged in series, one after the other, along a circuit—this means that as soon as one or more light bulbs fail, the circuit will break. Let \( T \) be the lifetime of the circuit—that is, the time until the circuit breaks.
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**(a)** What is the CDF of \( T \), the lifetime of the circuit?
**(b)** Next, suppose that we only check on the circuit once every second (assume the times \( T_1, T_2, T_3, T \) are measured in seconds). Let \( S \) be the first time we check the circuit and see that it’s broken. For example, if the circuit breaks after 3.55 seconds, we will only observe this when 4 seconds have passed, and so \( S = 4 \). Calculate the PMF of \( S \).
**(c)** Finally, suppose that instead of checking on the circuit every second, we instead do the following: after each second, we randomly decide whether to check on the circuit or not. With probability \( p \) we check, and with probability \( 1 - p \) we do not check. This decision is made independently at each time. Now let \( N \) be the number of times we check and see the circuit working. For example, if the circuit breaks at time 3.55, and our choices were to check at time 1 second, not to check at times 2 or 3 or 4, and to check at time 5, then \( N = 1 \), since the circuit was broken the 2nd time we checked.
What is the PMF of \( N \)? (Hint: start by finding the joint PMF of \( N \) and \( S \). It's fine if your answer is in summation form.)
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