I need help with this question on probability.?

Question

Choose independently two numbers a,b at random from the interval[0,1] with unifrom density. The point (a,b) is chosen at random in the unit square with uniform density. Find probability that
1. |a-b| < 0.5;
2. max(a,b) < 0.5;
3. min(a,b) < 0.5;
4. a < 0.5 and 1-b < 0.5;
5. conditions 1 and 4 both hold
6. a+b < 1/2

Any detailed explanations are greatly appreciated. Thank you!

Answer 1

1. |a-b|< 0.5 ---- P(b >=0.5 + a, a < 0.5) + P(a >= .5+b, b <0.5)
translated to P(b >0.5, a< 0.5) + P(a >0.5, b<0.5)
= 1/4 + 1/4 = 1/2
(detailed explanation - P(b >0.5) is 1/2, P(a<0.5) is 1/2 and P(a = 0.5) = P(b=0.5)= 0)

or this is better Y = |a-b| range of |a-b| goes from 0 to 1. P(Y <0.5 ) = 1/5

2. max(a,b) < 0.5 means P(a < 0.5) and P(b <0.5) = 1/2 x 1/2 = 1/4

3. min(a,b) < 0.5 means P(a < 0.5) or P(b <0.5) = 1/2 or 1/2 = 1/2.

4. a< 0.5 and 1-b< 0.5 means P(a<0.5) and P(b>0.5) = 1/2 x 1/2 = 1/4.

5. P(a< 0.5, b>0.5) = 1/2 x 1/2 = 1/4

6. a + b < 1/2 Let Y = a+ b. The range of a + b = 0 to 2.
P(Y < 1/2) = 1/4.

Answer 2

max of a: 1, min 0, max of b: 1, min 0, max of |a-b|: [0,1] too.
hence, i think number one, |a-b| should follow uniform density distribution also, which results in the probability of half.

2nd qn i don really understand, max(a,b) how many sets of (a,b) are u intending to extract, to attain a Max value? I think the qn should include the number of sets of (a,b) to be chosen first.. then can use binomial theorem to find the probability.

3rd qn too. Im not too sure anyway.

a<0.5 occurs half the time, so the probability is 0.5; 1-b<0.5=> b>0.5 which also occurs half the time. because the distribution is uniform. so combining both conditions should give a 1/4?

5 is v tricky, i dono too. the situation is: a<0.5; 0.5 a<0.5 is half; b
a+b <0.5 occurs 1/4 in probabily if the explanation follows the first qn.

Answer 3

Just graph your conditions on the xy plane. The area of the good part will be the probability you are looking for..