The real numbers x and y are chosen at random from the interval [0,1], as in x and y are greater than 0 and less than 1. Find the expected value of x+y.
Four points P, Q, R and S are chosen at random on the circumference of a circle. Find the probability that chord PQ and chord RS intersect.
Point Q is chosen at random inside equilateral triangle XYZ. Find the probability that Q is closer to the center of the triangle than to X, Y or Z. (In other words, let O be the center of the triangle. Find the probability that OQ is shorter than all of QX, QY and QZ)
Right triangle XYZ has legs of length XY=12 and YZ=6. Point D is chosen at random within the triangle XYZ. What is the probability that the area of triangle XYD is at most 12?
I really don't know how to even start on these. I'm really confused about the topic expected value.
"Right triangle XYZ has legs of length XY=12 and YZ=6. Point D is chosen at random within the triangle XYZ. What is the probability that the area of triangle XYD is at most 12?"
The probability is 2/5.
+128460 Point Q is chosen at random inside equilateral triangle XYZ. Find the probability that Q is closer to the center of the triangle than to X, Y or Z. (In other words, let O be the center of the triangle. Find the probability that OQ is shorter than all of QX, QY and QZ)
See the following image ;
We have an equilateral triangle with a side of 2 and a height of sqrt (3)
O is the point of intersection of the angle bisectors of the triangle and BY = BO = DO
So.....BY is 1/3 of the height of triangle XYZ
And triangle AYC is similar to triangle XYZ......so the area of triangle AYC = (1/3)^2 = (1/9) that of triangle XYZ
And Q will be closer to Y than it is to O when Q is located inside of triangle AYC
So.....using symmetry, we will have three such smaller triangles, so their total area = 3(1/9) = 1/3 that of triangle XYZ
So......the probability that Q will be closer to O than to any of the vertices of triangle XYZ = 2/3
