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**:\[\begin{array}{|c|c|}\hlinew & P(w) \\\hline0 & \\1 & \\2 & \\3 & \\4 & \\5 & \\6 & \\7 & \\\hline\end{array}\]**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 \( w \).

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)

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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