Because P(A n B) has been defined as x, and we calculated the value of P(A n B) = P(A) P(B) to be x² + 0.49x + 0.058, we now solve for x, the probability of both pumps failing. P(An B) = P(A) P(B) x = x² + 0.49x + 0.058 0 = x² - (C × x + 0.058 Solving the quadratic equation for x results in two values that represent the possible probabilities that the pumping. system will fail on any given day. Rounding both values to four decimal places, the lower value is x = the larger is x = and
Because P(A n B) has been defined as x, and we calculated the value of P(A n B) = P(A) P(B) to be x² + 0.49x + 0.058, we now solve for x, the probability of both pumps failing. P(An B) = P(A) P(B) x = x² + 0.49x + 0.058 0 = x² - (C × x + 0.058 Solving the quadratic equation for x results in two values that represent the possible probabilities that the pumping. system will fail on any given day. Rounding both values to four decimal places, the lower value is x = the larger is x = and
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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