The function f(x) = 7x for x = 1,2,4 and zero otherwise, is a probability mass function.
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- A joint probability mass function (pmf) is given by рх,у (—3, 0) — 0.18, Рx,у (-2, 1) — 0.08, рх,у (—1,4) — 0.16, рх,у (0,0) — 0.14, PX,Y (1, 2) = 0.04, рх,у (2, 1) — 0.4 The value of P(X >-1,Y > 1) isThe joint probability mass function of X and Y is given by p(1,1) = 0.35 p(1,2) = 0.05 p(1,3) = 0.05 p(2,1) = 0.05 p(2,2) = 0 Р(3, 1) — 0.1 p(2,3) = 0.1 p(3, 2) = 0.05 p(3,3) = 0.25 (a) Compute the conditional mass function of Y given X = 2: P(Y = 1|X = 2) P(Y = 2|X = 2) = P(Y = 3|X = 2) = (b) Are X and Y independent? (enter YES or NO)Determine the value of c that makes the function f(x, y) = c(x + y) a joint probability mass function over the nine points with x = 1,2,3 and y = 1,2,3. Give exact answer in form of fraction.
- Suppose that p(x, y), the joint probability mass function of X and Y, is given by p(1, 1) = 0.5, p(1,2)= 0.1, p(2, 1) = 0.1, p(2, 2) = 0.3 Calculate the conditional probability mass function of X given that Y = 1.The joint probability mass function of X and Y is given by p(1, 1) = 0.05 p(2, 1) = 0.1 p(1,2) = 0.05 p(1, 3) = 0.05 p(2,2) = 0 p(2, 3) = 0.05 p(3, 1) = 0.05 p(3,2) = 0.1 p(3, 3) = 0.55 (a) Compute the conditional mass function of Y given X = 2: P(Y = 1|X = 2) = ☐ P(Y = 2X = 2) = P(Y = 3|X = 2) (b) Are X and Y independent? (enter YES or NO) (c) Compute the following probabilities: P(X + Y > 2) = P(XY = 2) = P(¥ > 1) = |Consider the probability density fx (x) = a. eb x! where X is random variable whose allowable values range from x = -o to + o. Find (a) the CDF (b) the relation between a and b (c) the probability that x lies between 1 and 2.
- A shop receives a shipment of 1000 lamps. The probability that any individual lamp is defective is 0.2%. Assume the defectiveness is independent of cach lamp. Let X be the number of defective lamps in the batch of 1000. What is the probability mass function of X? O P(X = k) = (00) 0.002* (1 – 0.002)1000 , k = 0,1, 2, ..., 1000 O P(X = k) = 0.002* (1 – 0.002)1000–&, k = 0, 1, 2, .….., 1000 O P(X = k) = (1000) 0.002* (1 – 0.002)1000 , k = 1, 2, ..., 1000 O P(X = k) = (00)0.002100- (1 – 0.002)*, k = 0, 1, 2, ..., 1000Please show all stepsThe combined probability mass function of X and Y, p(x, y), as in the following data: p(1,1) =1/9, p(2,1)=1/3, p(3,1)=1/9, p(1,2)=1/9, p(2,2)=0, p(3,2)=1/18, p(1,3)=0, p(2,3)=1/6, p(3,3)=1/9 Calculate E[X|Y = i] for every i = 1, 2, 3
