Final answer:
The student's question is about finding the conditional probability P(0≤X≤4/11∧Y=4/5) for a point in a unit disk. The conditional nature of the probability reduces the problem to a simple uniform distribution between -1 and 1 for the X-coordinate given a fixed Y-coordinate of 4/5. The answer is obtained by considering the proportion of the specified range of X within the possible range given Y=4/5, resulting in a probability of 4/11.
Step-by-step explanation:
The student is asking for the conditional probability P(0≤X≤4/11∧Y=4/5) given a random point (X, Y) from a unit disk centered at the origin. To find this probability, we need to consider the fact that the value of Y is given as 4/5, which restricts the possible X-values within the unit disk. Based on the unit disk's radius, we know that the maximum and minimum values of X are 1 and -1, respectively. Hence, finding the probability that 0≤X≤4/11 is actually the same as finding the proportion of the length on the x-axis that lies between 0 and 4/11 given the fixed Y value of 4/5 within the disk.
Since the disk's radius is 1, and the fixed Y value is 4/5, by the Pythagorean theorem we can find the maximum X value that lies on the disk at this Y value, which would be the square root of (1 - (4/5)^2). However, as the maximum X value that the student is interested in is 4/11, which is less than 1, we do not need this calculation. The region of interest is a fraction of the disk's X-values at Y=4/5, and since the disk is symmetric about the X-axis, this is simply the X value range of interest (4/11) divided by the total possible range of X given Y=4/5. This results in a probability of 4/11, because for a fixed Y in a unit circle, the X values are uniformly distributed between the negative and positive square roots calculated above, and thus the length of 4/11 is merely a proportion of the total length.
The task involves basic knowledge of uniform distribution and conditional probability, as well as geometric interpretation within the context of a unit disk.