Each exam question posted to the study website Begg.com has a probability 0.6 of being answered correctly. However, for each question posted there is a probability 0.5 of being caught by the university authorities. What is the probability that a student who posts two questions receives at least one right answer and does not get caught on either question? You may assume that the events involved are mutually independent. I have tried but confused really on what method to use, idk if there is a formula for stuff like this I have tried 1-(0.4x0.5)2 which is 1 minus P(no answer x caught) twice but i got 0.96 which i think is wrong
Each exam question posted to the study website Begg.com has a probability 0.6 of being answered correctly. However, for each question posted there is a probability 0.5 of being caught by the university authorities. What is the probability that a student who posts two questions receives at least one right answer and does not get caught on either question? You may assume that the events involved are mutually independent. I have tried but confused really on what method to use, idk if there is a formula for stuff like this I have tried 1-(0.4x0.5)2 which is 1 minus P(no answer x caught) twice but i got 0.96 which i think is wrong
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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Each exam question posted to the study website Begg.com has a
I have tried but confused really on what method to use, idk if there is a formula for stuff like this
I have tried 1-(0.4x0.5)2 which is 1 minus P(no answer x caught) twice but i got 0.96 which i think is wrong
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Step 1: Given information
VIEWStep 2: Determine events
VIEWStep 3: Calculate the probability for the student receives at least one right answer.
VIEWStep 4: Calculate probability that the student does not get caught on either question
VIEWStep 5: Calculate probability of receiving at least one right answer and not getting caught either question
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