{"id":20321,"date":"2024-04-15T05:51:19","date_gmt":"2024-04-15T05:51:19","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20321"},"modified":"2024-04-15T05:51:19","modified_gmt":"2024-04-15T05:51:19","slug":"a-point-is-randomly-selected-with-uniform-probability-in-the-x-y-plane-within-the-rectangle-with-corners-at-0-0-1-0-1-2-and-0-2-if-p-is-the-length-of-the-position-vector-of-the-point","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-point-is-randomly-selected-with-uniform-probability-in-the-x-y-plane-within-the-rectangle-with-corners-at-0-0-1-0-1-2-and-0-2-if-p-is-the-length-of-the-position-vector-of-the-point\/","title":{"rendered":"A point is randomly selected with uniform probability in the X-Y. plane within the rectangle with corners at (0, 0), (1, 0), (1, 2) and (0, 2). If P is the length of the position vector of the point, the expected value of p2 is A. $$\\frac{2}{3}$$ B. 1 C. $$\\frac{4}{3}$$ D. $$\\frac{5}{3}$$"},"content":{"rendered":"

[amp_mcq option1=”$$\\frac{2}{3}$$” option2=”1″ option3=”$$\\frac{4}{3}$$” option4=”$$\\frac{5}{3}$$” correct=”option3″]<\/p>\n

The correct answer is $\\boxed{\\frac{4}{3}}$.<\/p>\n

The expected value of a random variable is the average of its possible values, weighted by their probabilities. In this case, the possible values of $P^2$ are $0, 1, 4, 9$, and their probabilities are $\\frac{1}{4}, \\frac{2}{4}, \\frac{1}{4}, \\frac{1}{4}$, respectively. Therefore, the expected value of $P^2$ is<\/p>\n

$$\\frac{0 \\cdot \\frac{1}{4} + 1 \\cdot \\frac{2}{4} + 4 \\cdot \\frac{1}{4} + 9 \\cdot \\frac{1}{4}}{4} = \\frac{4}{3}.$$<\/p>\n

Here is a more detailed explanation of each option:<\/p>\n