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- An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is hhh, then N (hhh) = = 3. Suppose that the random variable X is defined in terms of N as follows: X=6N-2N²-3. The values of X are given in the table below. Outcome hhh hth hht thh htt tth ttt tht Value of X-3 1 1 1 1 1 -3 1 Calculate the probabilities P (X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value X of X P(X=x) 0 0 0 00 XA QR code photographed in poor lighting, so that it can be difficult to distinguish black and white pixels. The gray color (X) in each pixel is therefore coded on a scale from 0 (white) to 100 (black). The true pixel value (without shadow) the code is Y = 0 for white, and Y = 1 for black. We treat X and Y as random variables. For the highlighted pixel in the figure is the gray color X = 20 and the true pixel value is white, i.e. Y = 0. We assume that QR codes are designed so that, on average, there are as many white as black pixels, which means that pY (0) = pY (1) = 1/2. In this situation, X is continuously distributed (0 ≤ X ≤ 100) and Y is discretely distributed, but we can still think about the simultaneous distribution of X and Y. We start by defining the conditional density of X, given the value of Y : fX|Y(x|0) = "Pixel is really white" fX|Y(x|1) =" Pixel is really balck " Use Bayes formula as given in the picture and find the probability for x = 20 like in the picture.3. Let X be the random variable that takes on the integers {0, 1, 2, ..., 15} with equal probabilities. Define a new random variable Y = X + A, where A is a random variable that takes on the values {-1, 0, 1} with equal probabilities. If the RVs X and A are independent, find the mutual information between X and Y.
- Q3. Let X and Y be two independent random variables, each representing the number of trials until the first success in a sequence of Bernoulli trials. Evaluate P(X = Y) assuming that the probability of success in a single trial(in case of X and Y both) is 0.4.An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (*) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is ttt, then N (ttt) = 0. Suppose that the random variable X is defined in terms of N as follows: X=2N -2. The values of X are given in the table below. Outcome ttt hth tht htt thh hhh hht tth Value of X -2 2 0 0 2 4 2 0 Calculate the probabilities P(X=*) of the probability distribution of X. First, fill in the first row with the valuesof X. Then fill in the appropriate probabilities in the second row. Value x of X ___ ___ ___ ___ P(x=x) ___ ___ ___ ___Let X,, X2, and X3 be independent random variables, each are binomially distributed with n = 100 and p = 0.2. Let A = X,– 2X2 and B = X3 + 3X1. Find PAB- %3D %3D
- A red die is rolled and then a green die. Let X=2(Spots on Red) + (Spots on Green). Determine the pmf for the random variable. In particular f(6)An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of tails in each outcome. For example, if the outcome is hth, then N (hth) = 1. Suppose that the random variable X is defined in terms of N as follows: X=2N² − 6N-1. The values of X are given in the table below. Outcome thh tth hhh hth ttt htt hht tht Value of X-5 -5 -1 -5 -1 -5 -5 -5 Calculate the probabilities P(X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value X of X P(X=x) 0 0 00 XEach of the random variables X and Y takes only 3 values {1,2,3} with the following probabilities: 1 1 0 1/6 1/6 y 2 1/6 0 1/6 3 1/6 1/6 0 2.
