This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Quiz 1t Started: October 5, 2006 11:21am Finished: October 5, 2006 11:40am Time spent: 18 min. 45 sec. Question 1 (5 points) If events A and B are not independent. What is the probability of their UNION? Student response: The probability equation for independent is P[AB] = P[A] P[B]. Since A and B are not independent then the probability of their union is zero. But if the event A and B are P[AB]= P[A] + P[B] then the probability is gt zero. Score: 2 / 5 Grader comments P(AuB) = P(A)+P(B)  P(AnB) Question 2 (5 points) Find the probability that a single toss of a fair die will result in the number 4 or less if it is given that the toss resulted in an even number. Student response: There is only one even number less than 4 and that is 2. Since there are 6 faces on a die then the probability of each number is 1/6, so the probability for the two would be 2/6 or 1/3. Score: 2 / 5 Grader comments NO this is a conditional probability =2/3 Total score: 4 / 10 = 40.0% Quiz 1m Started: October 6, 2006 4:58pm Finished: October 6, 2006 5:28pm Time spent: 30 min. 6 sec. Question 1 (5 points) A company makes computers which contain a chip which is supplied by three vendors with equal likelihood. Vendors are known to supply defective chips with the following probabilities 'A' .003; 'B' .002; and 'C' .001. If a computer is found to have a bad chip what is the probability that the chip came from vendor 'A'? Student response: Using Bayes Theorem P[A\B]=P[AB]P[B]/P[A] we find that the probability bad chip came from vendor A is (1/.002*.002/(.002+.001))/((1/.002)*(.002/(.002+.001)= .001 Score: 3 / 5 Grader comments math errorsP(bad)= 1/3(.001+.002+.003)=.002; P(Abad)= ((.003)1/3)/.002= 1/2 Question 2 (5 points) Given two events A and B, and that the probabilty of their intersection is the product of the individual probabilities. Are events A and B independent? Student response: If two events have a nonzero probability of occurrence, then by comparison it is established that two events cannot be independent. In order for them to be independent they must have an intersection. Since there is a probability of an intersection then events A and B are independent. Score: 5 / 5 Total score: 8 / 10 = 80.0% Quiz 1t 2004 Quiz 1m 2004 Question 1 (5 points) A production line manufactures 100 ohm resistors to a tolerance of 5%. The probability of any one resistor being out of tolerance is .02. If 5 resistors are slected at random what is the probability of exactly one being out of tolerance?...
View
Full
Document
This note was uploaded on 03/11/2008 for the course ECE 315 taught by Professor Clark during the Spring '08 term at Cal Poly Pomona.
 Spring '08
 Clark

Click to edit the document details