(8) X and Y are independent events. P(X)=0.3; P(Y)=0.4. Find(a) P(X and Y)(b) P(X | Y)(c) P(Y | X)(d) P(X or Y) (9) Repeat the previous problem if X and Y are mutually exclusive.

As per our guidelines we can solve first three sub part of first question and rest can be reposted.…

Answered: (8) X and Y are independent events.…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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An individual has a vNM utility function over money of u(x) = Vx, where x is final wealth. Assume the individual currently has $16. He is offered a lottery with three possible outcomes; he could gain an extra $9, lose $7, or not lose or gain anything. There is a 15% probability that he will win the extra $9. What probability, p, of losing $7 would make the individual indifferent between to play and to not play the lottery? (Make sure to answer in the form, 0.X, i.e. 0.25) Enter your answer here
Farmers face uncertainty over their earnings because rain is random. Let a farmer's yield be Y= 0 in the dry outcome and Y = 1 in the wet outcome. However, not all farmers are the same: Probability(Y= 1) = p1= 2/ 3 if type = 1 p0= 1/ 3 if type = 0 All farmers are risk averse, with utility U(Y) = √Y as their utility function. Farmers know their type. (a) Calculate each farmer's expected yield, E(Yi) =pI1 + (1- pi)0, expected utility ,E(Ui) =piU(1) + (1 - pi)U(0), and calculate the utility at E(Yi), U(E(Yi)), for i= 0, 1. Show these all on a graph with utility on the vertical axis and yield on the horizontal axis.
Could you double-check if my answers are correct for this question 2.17? If they are not correct, please point out the error and present the correction. Question. 2.17: a. P(A) + P(B) does not necessarily equal 1 because the events A and B could potentially overlap. In other words, it's possible for a request to be related to both the SPSS package and SAS simultaneously. The formula P(A) + P(B) - P(A ∩ B) gives the correct probability of either event A or event B occurring, taking into account the possibility of overlap. b. The complement of event A, denoted as A', is the event that the next request is not related to the SPSS package. The probability of the complement of A can be calculated as: P(A') = 1 - P(A) = 1 - 0.30 = 0.70 c. The union of events A and B, denoted as A ∪ B, is the event that the next request is related to either the SPSS package or SAS or both. The probability of the union of A and B can be calculated using the formula: P(A ∪ B) = P(A)…
(8) X and Y are independent events. P(X)=0.3; P(Y)=0.4. Find (a) P(X and Y) (b) P(X | Y) (c) P(Y | X) (d) P(X or Y)
Let (X1, X2, X3) be the random outcome of tossing a fair six-faced die three times, and let (Y(1), Y(2), Y(3) be the order statistics from (X1, X2, X3). We have Y(1) = min (X1, X2, X3), Y(3) = max (X1, X2, X3), and Y(1) < Y(2) < Y(3). Find Pr (Y(1) = 3).
Ex. A lottery ticket with two equally likely outcomes:(1) You win $2 (2) You win $0Let W be the amount won from the lottery. E[W] = 1VAR(W) =???
Possible values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with corresponding probabilities 0.15, 0.15, 0.35, and 0.35, respectively. (a) Calculate E(X) and then E(5 – X). E(X) = E(5 - X) (b) Would the repair facility be better off charging a flat fee of $95 or else the amount $ 150 ? [(5 - X)] Note: It is not generally true that E. E(Y) 150 The repair facility -Select--- v be better off charging a flat fee of $95 because E (5 - X)
Possible values of X, the number of components in a system submitted for repair that must be replaced, are 1, 2, 3, and 4 with corresponding probabilities 0.35, 0.15, 0.35, and 0.15, respectively. (a) Calculate E(X) and then E(5 - X). E(X) E(5-X) = (b) Would the repair facility be better off charging a flat fee of $70 or else the amount $ 150 (5-X) The repair facility ---Select--- be better off charging a flat fee of $70 because E ]= 150 (5-X)] ? Note: It is not generally true that : E ( =) = E(Y)
For mutually exclusive events R₁, R₂, and R₁, we have P(R₁)=0.05, P(R₂) = 0.3, and P(R₂) = 0.65. Also, P(QIR₁) = 0.4, P (Q1R₂) = 0.6, and P(Q|R3) =0.8. Find P (R3 | Q). P (R31Q) = (Type an integer or a simplified fraction.)
18a
(d) If X1,X2, ...,X, N(u, o²), and we use X as the point estimate u of the mean u, then (1) P{X = µ} = 1; (ii) P{X < A) = }: (iii) E(X) = µ; (iv) in approximately 95% of samples drawn, X will equal µ.
(b) Let X~N(40,144) and if Y = 2X – 1 Find the following probabilities: (i) P(X 50) (iii)P(35 < X < 45),(iv)P(-45 < Y < 45).(v)P(-50 < Y < 100)

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(8) X and Y are independent events. P(X)=0.3; P(Y)=0.4. Find (a) P(X and Y) (b) P(X | Y) (c) P(Y | X)