CH5
... A discrete random variable has either a finite or a countable number of values. This chapter deals with discrete random variables. A continuous random variable has infinitely many values, and those values can be associated with measurements on a continuous scale in such a way that there are no gaps ...
... A discrete random variable has either a finite or a countable number of values. This chapter deals with discrete random variables. A continuous random variable has infinitely many values, and those values can be associated with measurements on a continuous scale in such a way that there are no gaps ...
Chapter 2 Probability
... Because of this symmetry, we then say that A and B are independent. From the definition of either P (A|B) or P (B|A), it follows then that P (A∩B) = P (A)P (B). Two events A1 and A2 are said to be independent (statistically or in the probability sense), if P (A1 ∩A2 ) = P (A1 )P (A2 ). When P (A1 ∩A ...
... Because of this symmetry, we then say that A and B are independent. From the definition of either P (A|B) or P (B|A), it follows then that P (A∩B) = P (A)P (B). Two events A1 and A2 are said to be independent (statistically or in the probability sense), if P (A1 ∩A2 ) = P (A1 )P (A2 ). When P (A1 ∩A ...
Question 1:
... a) Investigate the independence of A and B using the product of their probabilities. b) Investigate the independence of A and C using P(C/A). c) Investigate the independence of B and C using P(B/C). d) Can you find the sum of the probabilities of all simple events outside A B C ? Question 6: Of ...
... a) Investigate the independence of A and B using the product of their probabilities. b) Investigate the independence of A and C using P(C/A). c) Investigate the independence of B and C using P(B/C). d) Can you find the sum of the probabilities of all simple events outside A B C ? Question 6: Of ...
Probability Distributions An Example With Dice The Finite Uniform
... Thus, X + Y has distribution Bn+m,p. An easier argument: Perform n + m Bernoulli trials. Let X be the number of successes in the first n and let Y be the number of successes in the last m. X has distribution Bn,p, Y has distribution Bm,p, X and Y are independent, and X + Y is the number of successes ...
... Thus, X + Y has distribution Bn+m,p. An easier argument: Perform n + m Bernoulli trials. Let X be the number of successes in the first n and let Y be the number of successes in the last m. X has distribution Bn,p, Y has distribution Bm,p, X and Y are independent, and X + Y is the number of successes ...
MATH 461/661 Homework 4 Solutions
... 3. 3.2.4 Let p ≡ P (audit) = 0.153 for each of the n = 6 corporations. Then P (≥ 2 audited) should follow the binomial distribution with parameters n and p. That implies P (≥ 2 audited) = 1 − P (0 or 1 audited) = 1 − (1 − p)6 − 6p(1 − p)5 ≈ 0.23 4. Question: Consider the experiment of rolling six si ...
... 3. 3.2.4 Let p ≡ P (audit) = 0.153 for each of the n = 6 corporations. Then P (≥ 2 audited) should follow the binomial distribution with parameters n and p. That implies P (≥ 2 audited) = 1 − P (0 or 1 audited) = 1 − (1 − p)6 − 6p(1 − p)5 ≈ 0.23 4. Question: Consider the experiment of rolling six si ...
Lecture 21 - WordPress.com
... Next, we discuss the concept of INDEPENDENT EVENTS: INDEPENDENT EVENTS: Two events A and B in the same sample space S, are defined to be independent (or statistically independent) if the probability that one event occurs, is not affected by whether the other event has or has not occurred, that is P( ...
... Next, we discuss the concept of INDEPENDENT EVENTS: INDEPENDENT EVENTS: Two events A and B in the same sample space S, are defined to be independent (or statistically independent) if the probability that one event occurs, is not affected by whether the other event has or has not occurred, that is P( ...
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... Description and rationale: This course deals with the basic discussion of experimental design and the practical use of statistical procedures. It is important to design meaningful experiments for establishing causal relationship. Also, it is necessary to study the principles of probability to evalua ...
... Description and rationale: This course deals with the basic discussion of experimental design and the practical use of statistical procedures. It is important to design meaningful experiments for establishing causal relationship. Also, it is necessary to study the principles of probability to evalua ...
Handout1B - Harvard Math Department
... b) Write down the subsets of your sample space that correspond to the event that outcome 1 occurs in the second experiment. c) Suppose that we have a theoretical model of the situation that predicts equal probability for any of the three outcomes for any one given experiment. Our model also says tha ...
... b) Write down the subsets of your sample space that correspond to the event that outcome 1 occurs in the second experiment. c) Suppose that we have a theoretical model of the situation that predicts equal probability for any of the three outcomes for any one given experiment. Our model also says tha ...
P - OSU Physics
... So, if we throw darts at random at our rectangle then the probability () of a dart landing inside the circle is just the ratio of the two areas, p/4. The we can determine p using: The error in p is related to the error in by: ...
... So, if we throw darts at random at our rectangle then the probability () of a dart landing inside the circle is just the ratio of the two areas, p/4. The we can determine p using: The error in p is related to the error in by: ...
Making Statistical Connections in Middle School
... Making Statistical Connections in Middle School ...
... Making Statistical Connections in Middle School ...
Conditional probability and independent events
... given Y naturally forms a matrix; let's call it q where qij is given by (11). Suppose, as in section 1.2.4, we represent the joint pmf by means of the matrix p where pij = Pr{X = xi, Y = yj}. Then the individual pmf of X is given by the row sums of p. So the matrix q for the conditional pmf of Y giv ...
... given Y naturally forms a matrix; let's call it q where qij is given by (11). Suppose, as in section 1.2.4, we represent the joint pmf by means of the matrix p where pij = Pr{X = xi, Y = yj}. Then the individual pmf of X is given by the row sums of p. So the matrix q for the conditional pmf of Y giv ...
