a rectangle is drawn with its adjacent sides chosen from independent and identically distributed uniform distributions over the interval (0,1) . What is the probability that the area of the rectangle is more than 0.5

[tex]\displaystyle P(XY > 0.5) = \int_{0.5}^1 \int_{0.5/y}^1 dx \, dy \\\\ = \int_{0.5}^1 \left(1-\frac1{2y}\right) \, dy \\\\ = \left(1 - \frac12 \ln(1)\right) - \left(\frac12 - \frac12 \ln\left(\frac12\right)\right) \\\\ = \boxed{\frac{1-\ln(2)}2}[/tex]

or approximately 0.1534.

facundo3141592 facundo3141592 · Answer 2 · 2021-11-24T11:32:22+01:00

We want to find the probability that the area of the rectangle is more than 0.5, we will see that the probability is P ≈ 0.1534

We know that the area of a rectangle of length L and width W is given by:

A = L*W

In this case, if we know that:

L can be any value between (0, 1)
W can be any value between (0, 1)

And we know that the probabilities are equally distributed.

So we impose:

L*W > 0.5

Then we define a lower limit for W as a function of L as:

W = 0.5/L

Then we just compute the lower bound (which is an approximation) of the probability by integrating:

[tex]P = \int\limits^1_{0.5} \int\limits^1_{0.5/L} dLdW[/tex]

We can rewrite this as:

[tex]P = \int\limits^1_{0.5} ({1 - \frac{1}{2L} } )\, dL[/tex]

[tex]P = (1 - 0.5) - (1/2)*(Ln(1) - Ln(0.5))\\\\P = 0.5 - (1/2)*(Ln(1/0.5))\\\\P = 0.5 - (1/2)*Ln(2) = 0.5*(1 - ln(2)) = 0.1534[/tex]

If you want to learn more, you can read:

https://brainly.com/question/23044118