1. ## find the probabilities

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

A: The magnitude of the sum of the two numbers is greater than 2/3
B: At lest one of the numbers is greater than 2/3
C: The two numbers are equal
D: Alice's number is greater than 2/3

Find the following probabilities:

a) Pr(A)
b) Pr(B)
c) Pr(A∩B)
d) Pr(C)
e) Pr(A∩D)

ok for each part i guess i draw up a venn diagram? but i'm not entirely sure because the question is worded differently from the questions i've had before.

Could anyone help me understand the line "We assume a uniform probability law under which the probability law under which the probability of an event is proportional to its area."

Not looking for a cheat just a way to start it.

2. I will help you with part b. It will be a model for the rest.
In the unit square below, a point in the area shaded yellow will have at least one of its coordinates greater than $\displaystyle \frac{2}{3}$.
Therefore the probability of event $\displaystyle B$ is just that area divided by the total area.

Then you model the other events.

3. thanks for your help its much appreciated. Before i move on would i be right in saying Pr(B) = 5/9 ?

4. Originally Posted by djmccabie
thanks for your help its much appreciated. Before i move on would i be right in saying Pr(B) = 5/9 ?
Correct.

5. for some reason i just cant remember how to do Pr(sum of 2 numbers)>(a number). I'm thinking it should be Pr(alice∩Bob)>5/9 but this cant be right because i don't know the values for alice or bob :/ do i need to use binomial?

6. Originally Posted by djmccabie
for some reason i just cant remember how to do Pr(sum of 2 numbers)>(a number). I'm thinking it should be Pr(alice∩Bob)>5/9 but this cant be right because i don't know the values for alice or bob :/ do i need to use binomial?
Graph the line $\displaystyle a+b>\frac{2}{3}$.
Shade the correct area inside the square.

7. I am trying to solve the same problem.
I found some answers but I do not know if I am correct.

a) 17/18
b) 5/9
c) 5/36
d) 1/2
e) 1/12

Some of the answers are strange. Can anyone tell me if the answers are correct and if not, what answers are incorrect to think them again please?

You will see the your part (a) in not correct.

As for part (c) the answer will suprise you. Think in terms of area.

9. I'm just not getting this

10. is part a 7/9?

11. also could (part b) = (part c)?

12. Originally Posted by djmccabie
I'm just not getting this
Here is another try.
The blue line is $\displaystyle b=-a+\frac{2}{3}$
Event (A) is the yellow shaded area: $\displaystyle a+b>\frac{2}{3}$.
That area is the probability.

Note that event (B) is a subevent of (A).
Thus $\displaystyle A\cap B=B$.

13. Originally Posted by djmccabie
also could (part b) = (part c)?
Is there any area in event (D)?

14. Originally Posted by Plato
Note that event (B) is a subevent of (A).
Thus $\displaystyle A\cap B=B$.

doesn't this show that part(b) = part(c)

ie Pr(B) = Pr(A∩B)

I plotted the square the same as you did also and came out with 7/9 is this wrong?

15. Originally Posted by djmccabie
doesn't this show that part(b) = part(c)
ie Pr(B) = Pr(A∩B)
I plotted the square the same as you did also and came out with 7/9 is this wrong?
In the OP $\displaystyle D=A\cap B$ so $\displaystyle P(D)=P(B)$.
$\displaystyle C$ the two numbers are equal.