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- .I already splited the questions ,? now pls answer it all i will rate you helpful i dont care about bartleby rulesSuppose the probability of having blood type A is 0.10. Let X be the number of people with type A in the experiment of selecting one thousand random people. 1. The values of the random variable X is x = _____. A. {1, 2, …} B. {0, 1, 2, …} C. {1, 2, …, 1000} D. {0, 1, 2, …, 1000} 2. Which of the following is(are) correct? Then, explain whyI. The probability of each value of x is between 0 & 1II. The probability that one thousand people will have blood type B is 1 A. I onlyB. II onlyC. Both I and IID. Neither I nor IIQ5A A box contains 68 red balls, 80 white balls, and 55 blue balls. (a) If one ball is drawn from that box at random, determine the probability that it is not white. (b) If three balls are drawn at random, find the probability that they are drawn in the order blue, white, and red (each ball is not replaced). U..
- Consider two random variables X and Y with p(X, Y) = 0. Which of the following statements is always correct? O a. X and Y are independent but correlated. O b. X and Y are uncorrelated. O c. X and Y are independent. O d. X and Y are uncorrelated but not independent.in which three balls are selected simultaneously and ruI 4 II yu. randomly from a box. But assume that the box contains 8 red balls and 4 green balls. Also assume that each red ball selected pays $0.60, that each green ball selected pays $0.30, and that it costs $1.10 to play the game. A random variable X is defined to be the net payoff for a player. How many different values are possible for the random variable X? 4 Fill in the table below to complete the probability density function. Be certain to list the values of X in ascending order. Probability Value of X -0.20 0.10 0.40 0.70Three balls are selected from a bag that contains 3 blue, 2 green, and 5 yellow balls. Here, we define random variables as B = number of blue balls drawn, G = number of green balls drawn, Y = number of yellow balls drawn. Please list the following values:
- A 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.In which three balls are selected simultaneously and randomly from a box. But assume that the box contains 8 red balls and 4 green balls. Also assume that each red ball selected pays $0.60, that each green ball selected pays $0.30, and that it costs $1.10 to play the game. A random variable X is defined to be the net payoff for a player. How many different values are possible for the random variable X?An insurance company covers 100$ per day for up to three days staying in a hospital, and 50$ per day for any day after the first three days if any. The number of days the patients used to stay in the hospital is a random variable with PMF: 6-k k = 1, 2, 3, 4, 5 Px(k) =. 15 else the expected amount the company will pay is: Select one: а. 220 b. 230 c. 166.67 d. 233.33 е. 200
- Determine which of these statements is correct/incorrect and explain the reason why. a. X is a discrete random variable. Therefore, P(9 < X < 11) = P(X=9) + P(X=10) + P(X=11) b. The probability of flipping two coins will result in one head and one tail is 1/2 c. K and L are mutually exclusive events. P(K)=0.3 and P(L)=0.4. Hence, P(K∩L) = 0.12 d. There are mass points that have a probability of zero. That is, if x is a mass point, then it's possible that P(X=x) = zero.M3How was the correct answer of 0.0741 attained?
