Probability of Events A, B, C, D in a Uniform Probability Law

[nacho-man]

Alice and bob each choose a number at rndom between zero and two. We assume a uniform probability law under which the probability of an event is proportional to it sarea. Consider the following events:

A: The magnitude of the difference of the two numbers is greater than 1/3
B: At least one of the numbers is greater than 1/3
C: The two numbers are equal
D: Alice's number is greater than 1/3

So, find $ P(A), P(B), P(A\cap B) , P(C), P(D), P(A\cap D)$

 

[chisigma]

Alice and bob each choose a number at rndom between zero and two. We assume a uniform probability law under which the probability of an event is proportional to it sarea. Consider the following events:

A: The magnitude of the difference of the two numbers is greater than 1/3
B: At least one of the numbers is greater than 1/3
C: The two numbers are equal
D: Alice's number is greater than 1/3

So, find $ P(A), P(B), P(A\cap B) , P(C), P(D), P(A\cap D)$

Let's give answers in increasing order of difficulty...

c) indicating with X and Y the two numbers the joint p.d.f. is...

$\displaystyle f (x,y) = \frac{1}{4}, 0 < x < 2,\ 0 < y < 2\ (1)$

... and 0 otherwise. The probability that X and Y are equal is the area of the segment y=x in the x,y plane... and the area of a segment is 0...

d) the marginal p.d.f. of X [the Alice's number...] is...

$\displaystyle f(x) = \frac{1}{2}, 0 < x < 2\ (2)$

... and 0 otherwise, so that...

$\displaystyle P \{ X > \frac{1}{3} \} = \int_{\frac{1}{3}}^{2} \frac{d x}{2} = \frac{5}{6}\ (3)$

b) taking into account what is written in c) we have that the requested probability is...

$\displaystyle P = 1 - \int_{0}^{\frac{1}{3}} \int_{0}^{\frac{1}{3}} \frac{d x\ d y}{4} = \frac{35}{36}\ (4)$

The point a) requires a little more effort and it will be attached in next post...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

Alice and Bob each choose a number at random between zero and two. We assume a uniform probability law under which the probability of an event is proportional to its area. Consider the following events:

A: The magnitude of the difference of the two numbers is greater than 1/3
B: At least one of the numbers is greater than 1/3
C: The two numbers are equal
D: Alice's number is greater than 1/3

So, find $ P(A), P(B), P(A\cap B) , P(C), P(D), P(A\cap D)$

View attachment 1526​

The question suggests representing the probability of an event by an area. So take Alice's number to be the $x$-coordinate and Bob's number to be the $y$-coordinate, in a $2\times2$ square, as in the diagram. For part A, you want $|x-y|>\frac13$, so you must avoid the region shaded yellow in the diagram. The required probability is then the fraction of the square that is not shaded yellow. So it is the combined area of the two unshaded triangles, divided by the area of the whole square. The other parts of the question can be done in a similar way.