Week5
... that the corporations intranet has been suffering with a number of faults and is unavailable much of the time. He arranges for the collection of data which details the number of times where the intranet has been down, and for how long, over the last three hundred business days. The results are given ...
... that the corporations intranet has been suffering with a number of faults and is unavailable much of the time. He arranges for the collection of data which details the number of times where the intranet has been down, and for how long, over the last three hundred business days. The results are given ...
投影片 1
... • Give an example of a single point in the sample space? • How might you depict the full sample space? • What is an example of an “event”? ...
... • Give an example of a single point in the sample space? • How might you depict the full sample space? • What is an example of an “event”? ...
Statistics and Probability
... Probability distribution of a discrete variable becomes more normal as sample size increases. Learning Competencies Students will be able to define a random variable, assign probabilities to its sample space, and graph the distribution of the random variable. calculate and interpret in context t ...
... Probability distribution of a discrete variable becomes more normal as sample size increases. Learning Competencies Students will be able to define a random variable, assign probabilities to its sample space, and graph the distribution of the random variable. calculate and interpret in context t ...
Lecture 5.1
... Recall from section 2.5: Two events A and B are independent when P ( A ∩ B ) = P ( A) ∗ P (B ) . If we remember that p(x, y) = P(X = x, Y = y) should be interpreted as p ( x, y ) = P ( X = x and Y = y ) = P ( X = x ∩ Y = y ) then it makes sense to state the following definition: Two discrete random ...
... Recall from section 2.5: Two events A and B are independent when P ( A ∩ B ) = P ( A) ∗ P (B ) . If we remember that p(x, y) = P(X = x, Y = y) should be interpreted as p ( x, y ) = P ( X = x and Y = y ) = P ( X = x ∩ Y = y ) then it makes sense to state the following definition: Two discrete random ...
Name - Garnet Valley School
... e. In how many different ways could they select a pitcher, a catcher, and a first baseman? You Try 1. A card is drawn at random from a normal 52-card deck. a. What name is given to the act of drawing the card? ...
... e. In how many different ways could they select a pitcher, a catcher, and a first baseman? You Try 1. A card is drawn at random from a normal 52-card deck. a. What name is given to the act of drawing the card? ...
Probability:
... a definite outcome. Usually the outcome is in the form of a description, count, or measurement. For example: If you toss a coin, there are only 2 possible outcomes (heads or tails). Sample Space - set of all possible outcomes. It is especially convenient to know the sample space where all outcomes a ...
... a definite outcome. Usually the outcome is in the form of a description, count, or measurement. For example: If you toss a coin, there are only 2 possible outcomes (heads or tails). Sample Space - set of all possible outcomes. It is especially convenient to know the sample space where all outcomes a ...
Slide 1
... • Suppose we have an unfair coin and wish to estimate the outcome (head or tail) from observing a series of coin tosses. • q = probability of tossing a head. ...
... • Suppose we have an unfair coin and wish to estimate the outcome (head or tail) from observing a series of coin tosses. • q = probability of tossing a head. ...
PPT - School of Computer Science
... In any one game, each is equally likely to win What is most likely length of a “best of 7” series? Flip coins until either 4 heads or 4 tails Is this more likely to take 6 or 7 flips? ...
... In any one game, each is equally likely to win What is most likely length of a “best of 7” series? Flip coins until either 4 heads or 4 tails Is this more likely to take 6 or 7 flips? ...
Equally Likely outcomes
... Coincidence: The Birthday Problem The event E c corresponds to the set of lists where all birthdays on the list are different.. By the multiplication principle there are 365 · 364 · 363 · 362 · · · 345 = P(365, 20) such lists (a 51-digit number). Now P(E c ) = P(365, 20)/36520 ≈ 0.588 and hence P(E ...
... Coincidence: The Birthday Problem The event E c corresponds to the set of lists where all birthdays on the list are different.. By the multiplication principle there are 365 · 364 · 363 · 362 · · · 345 = P(365, 20) such lists (a 51-digit number). Now P(E c ) = P(365, 20)/36520 ≈ 0.588 and hence P(E ...