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Suppose that Y1 is a binomial random variable with four trials and success probability 0.7 and that Y2 is an independent binomial random variable with three trials and success probability 0.5. Let W = Y1 + Y2. W does not have a binomial distribution. Find the probability mass function for W. [HINT: P(W = 0) = P(Y1 = 0, Y2 = 0); P(W = 1) = P(Y1 = 1, Y2 = 0) + P(Y1 = 0, Y2 = 1); etc.] (Round your answers to four decimal places.)

Suppose that Y1 is a binomial random variable with four trials and success probability 0.7 and that Y2 is an independent binomial random variable with three trials and success probability 0.5. Let W = Y1 + Y2. W does not have a binomial distribution. Find the probability mass function for W. [HINT: P(W = 0) = P(Y1 = 0, Y2 = 0); P(W = 1) = P(Y1 = 1, Y2 = 0) + P(Y1 = 0, Y2 = 1); etc.] (Round your answers to four decimal places.)

A First Course in Probability (10th Edition)
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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Suppose that 

Y1

 is a binomial random variable with four trials and success probability 0.7 and that 

Y2

 is an independent binomial random variable with three trials and success probability 0.5. Let 

W = Y1 + Y2.

 W does not have a binomial distribution. Find the probability mass function for W. [HINT: 

P(W = 0) = P(Y1 = 0, Y2 = 0);

 

P(W = 1) = P(Y1 = 1, Y2 = 0) + P(Y1 = 0, Y2 = 1);

 etc.] (Round your answers to four decimal places.)

**Problem Overview**: Suppose that \( Y_1 \) is a binomial random variable with four trials and a success probability of 0.7, and \( Y_2 \) is an independent binomial random variable with three trials and a success probability of 0.5. Let \( W = Y_1 + Y_2 \). \( W \) does not follow a binomial distribution. The task is to find the probability mass function for \( W \). **Instructions & Hint**: - You need to calculate probabilities for different values of \( W \), such as: - \( P(W = 0) = P(Y_1 = 0, Y_2 = 0) \) - \( P(W = 1) = P(Y_1 = 1, Y_2 = 0) + P(Y_1 = 0, Y_2 = 1) \) - Continue this pattern for other values of \( W \). **Table**: \[ \begin{array}{|c|c|} \hline w & P(w) \\ \hline 0 & \\ 1 & \\ 2 & \\ 3 & \\ 4 & \\ 5 & \\ 6 & \\ 7 & \\ \hline \end{array} \] **Details**: - Each \( w \) represents a possible value of \( W \). - The corresponding \( P(w) \) is the probability of \( W \) equaling that value. - Answers should be rounded to four decimal places. **Tools**: - You might need to use technological aids (like a calculator or software) to compute these probabilities accurately. Consider breaking down the probabilities using combinations where necessary, and remember to consider all combinations of successes for \( Y_1 \) and \( Y_2 \) that result in each \( w \).
Transcribed Image Text:**Problem Overview**: Suppose that \( Y_1 \) is a binomial random variable with four trials and a success probability of 0.7, and \( Y_2 \) is an independent binomial random variable with three trials and a success probability of 0.5. Let \( W = Y_1 + Y_2 \). \( W \) does not follow a binomial distribution. The task is to find the probability mass function for \( W \). **Instructions & Hint**: - You need to calculate probabilities for different values of \( W \), such as: - \( P(W = 0) = P(Y_1 = 0, Y_2 = 0) \) - \( P(W = 1) = P(Y_1 = 1, Y_2 = 0) + P(Y_1 = 0, Y_2 = 1) \) - Continue this pattern for other values of \( W \). **Table**: \[ \begin{array}{|c|c|} \hline w & P(w) \\ \hline 0 & \\ 1 & \\ 2 & \\ 3 & \\ 4 & \\ 5 & \\ 6 & \\ 7 & \\ \hline \end{array} \] **Details**: - Each \( w \) represents a possible value of \( W \). - The corresponding \( P(w) \) is the probability of \( W \) equaling that value. - Answers should be rounded to four decimal places. **Tools**: - You might need to use technological aids (like a calculator or software) to compute these probabilities accurately. Consider breaking down the probabilities using combinations where necessary, and remember to consider all combinations of successes for \( Y_1 \) and \( Y_2 \) that result in each \( w \).
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