**Probability and Independence of Events**A student stays at home. Let event N = the student watches Netflix. Let event Y = the student watches the very educational YouTube videos made by her/his instructor.Suppose P(N) = 0.1, P(Y) = 0.8, and P(N and Y) = 0.08.**Are N and Y independent?**- Yes, they are independent.- No, they are not independent.- It can't be decided with given info.- No answer text provided.

Answered: A student stays at home. Let event N=…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Picture an experiment that observes the # of computers in use at a five-computer lab. Let A, B be the events defined as A = {3,4,5} and B={1,3}. Using set notation (i.e. using {}), the event (A union B)^C where superscript denotes the complement, would be equal to:
At the close of the stock markets on a day in 2017, there were 6100 companies listed with certain stock exchanges. On this day, the probabilities of advancing, declining, and unchanged company stocks appears in the table for the two exchanges. Find the probability that a stock was listed on Exchange 2 given that it advanced. Percentage of Probability of Probability of Companies Probability the Stock Stock Stocks Remain Unchanged Exchange Exchange 1 Exchange 2 Listed Advancing Declining 0.2466 0.5052 0.7237 0.0297 0.4948 0.6009 0.3455 0.0536 ...... Let F be Exchange 2 and E be an advance. P(F) = and P(F') =D
a statistic quiz has 10 multiple choice questions, whre there are four incorrect options and one correct option to choose from if someone didnt study for the quiz and is randomly guessing on each of the 10 questions what is the probability that a person who didn't study and is guessing get at least 7 questions correcct
Hi i’m in grade 12 Data Management and i need help with this practice question please use « let » statements and permutations if needed to solve. use probability formulas because i’m not sure how to apply them.
Suzette does an experiment where she collects date from 5 women snd obtains the scores: 4, 4, 5, 2, 5. She hypothesizes that the population mean for the population from which these scores are drawn is 3. She wants to test whether this hypothesis is true. What is the alternative hypothesis she is testing?
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At the close of the stock markets on a day in 2017, there were 6100 companies listed with certain stock exchanges. On this day, the probabilities of advancing, declining, and unchanged company stocks appears in the table for the two exchanges. Find the probability that a stock was listed on Exchange 2 given that it advanced. Percentage of Probability of Probability of Companies Listed Probability the Stock Stock Stocks Remain Unchanged Exchange Exchange 1 Exchange 2 Advancing Declining 0.5052 0.7237 0.2466 0.0297 0.4948 0.6009 0.3455 0.0536 Let F be Exchange 2 and E be an advance. P(F) = .4948 and P (F') = .5052 The probability that a stock was listed on Exchange 2 given that it advanced is (Do not round until the final answer. Then round to four decimal places as needed.)
Art Pixel - Probability of Compo X Fatima Khan - Gall-Student_Inc X Fatima Khan - Gall- Studen Cs.google.com/spreadsheets/d/1dWslroQAl915ikzODL8Fg6r6Fyyzb9nOrA-CEEKWSU8/edit#gid3881570761 M Minooka Linksv first day invite - Go... -- Gall - Student_Ind and Dependent Probability Pixel Art D O Insert Format Data Tools Add-ons Help Last edit was seconds ago 100% .00 123 Default (Ari.. BISA 田 ,州。 10 D. 5. In a bag there are 2 red marbles, 3 white marbles and 5 blue marbles. Once a marble is selected, it is NOT replaced. Find the P(Blue then Red). Answer as percent.
Look at the 4 probability rules beow , an provide a REAL LIFE example for each. Also, with a Real LIfe EXAMPLE, explain how Disjoint events CAN be dependent. Rule 1. The probability P(A) of any event A satisfies 0 ≤ P(A) ≤ 1. Rule 2. If S is the sample space in a probability model, then P(S) = 1. Rule 3. Two events A and B are disjoint if they have no outcomes in common and so can never occur together. If A and B are disjoint, P(A or B) = P(A) + P(B) This is the addition rule for disjoint events. Rule 4. For any event A, P(A does not occur) = 1 − P(A)
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signment 2 docs.google.com/forms/d/e/1FAIpQLScmbOKQnvrrgoopk9LLokexf1ZnWZY1CBx6BuNKguG4 k=9/2 Question * Suppose that 70% of all diners choose vegetarian meals when eating away from home. On a randomly chosen day there are 20 diners eating in the restaurant. The probability of having more than m vegetarian diners is 0.7723. Then: m=13 m=15 m=12 None of these Question Let XE (0, 1) and YE (0, 1) be two independent binary random variables. If P(X= 0) = 0.3 and P(Y = 0) = 0.7, then P(X+Y≥ 1) is equal to: 0.84 0.79 O 0.76 | 0.86 B Very high UV GO
Probability and statistics. Please be very fast

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