PROBABILITY
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“Learn all about probability, the branch of mathematics that deals with the likelihood of events occurring. Explore fundamental concepts like outcomes, events, permutations, combinations, and probability distributions. Perfect for students and professionals looking to understand randomness, make data-driven decisions, and apply probability in real-life scenarios and statistical analysis.”
Mastering Probability Theory: A Comprehensive Guide to Random Variable
Probability theory is a fascinating subject that has many applications in the real world. Understanding the basics of random variables and probability distributions is essential for anyone working in a field that deals with uncertainty. By mastering probability theory, you can make better decisions and improve your ability to analyze and interpret data. READ MORE...
Practical Examples of Continuous Random Variables
Practical illustrations of Random Variables that we are exposed to in our daily life READ MORE...
Probability Cause and Effect Problem
Question What does it mean for one event ? to cause another event ? - for example, smoking (?) to cause cancer (?)? There is a long history in philosophy, statistics, and the sciences of trying to clearly analyze the concept of a cause. One tradition says that causes raise the probability of their effects; we may write this symbolically as \( ?(?|?) > ?(?) \) - - - - - - - - - - (1) a) Does equation (1) imply that ?(?|?) > ?(?)? If so, prove it. If not, give a counter-example. b) Another way to formulate a probabilistic theory of causation is to say that \( P (E | C) > P(E | C^c) \) […] READ MORE...
Probability Problem: Suppose you roll a fair die two times. Let ? be the event “THE SUM OF THE THROWS EQUALS 5” and ? be the event “AT LEAST ONE OF THE THROWS IS A 4”. Solve for the probability that the sum of the throws equals 5, given that at least one of the throws is a 4. That is, solve ?(?|?).
Solution We have A = (1,4), (2,3), (3,2), (4,1) B = (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3), (4,5), (4,6) \( P(A|B) = \dfrac {P(A∩B)}{P(B)} \) \( A∩B = (1,4), (4,1) \) The sample space comprises of 6×6 = 36 eventsHence,\( P(A∩B) = \dfrac{2}{36} = \dfrac{1}{18} \)\( P(B) = \dfrac{11}{36} \) Thus, \( P(A|B) = \dfrac {\dfrac{2}{36} } { \dfrac{11}{36} } \) or \( P(A|B) = \dfrac{2}{11} \) (Required probability) READ MORE...