[Ezra Halleck]

4. Suppose you roll two fair dice. Let A= doubles, B= sum is 6
(a) Are A and B disjoint?
(b) Determine P(A), P(B) and P(A and B).
(c) Are A and B independent?

[dannyCR09]

4.
a) A and B are disjoint events.

b) P(A) = 6/36 or 1/6
P(B) = 2/36 or 1/18
P(A & B) = 8/36 or 2/9

c) No, A and B are not independent.
P(A ∩ B) = P(A) * P(B)
0 = 6/36 * 2/36
0 = 1/108
X

[Ezra Halleck]

I suggest that you start off by listing exactly what A and B are. Then you can perhaps begin answering the questions. To be blunt, you write P(A & B) = 8/36 or 2/9 but earlier, you said that they were disjoint. It would be helpful to have definitions, since we have not seen this material for some time.

[dannyCR09]

When you say P(A and B), do you mean the union or the intersection?

[Ezra Halleck]

and=intersection
or=union

[dannyCR09]

Roll of two fair dice
P(A) = doubles , P(B) = even sum of 6

4.
a) A and B are disjoint events.
Disjoint – when two events cannot occur at the same time

b) The probability of P(A) happening – getting a double when you roll the two dice (1,1; 2,2; 3,3, etc.) = 6/36 or 1/6
The probability ofP(B) happening – getting two even numbers to add up to 6 (4,2; 2,4) = 2/36 or 1/18
The intersection of P(A & B) is 0 because A and B cannot occur at the same time.

c) A is independent of B if P(A ∩ B) = P(A) * P(B) and vice versa
No, A and B are not independent.
P(A ∩ B) = P(A) * P(B)
0 = 6/36 * 2/36
0 = 1/108
X

[Ezra Halleck]

I see where the problem is now. There are more ways to get a sum of 6, e.g., 1,5.

[dannyCR09]

So, when u put “B= even sum is 6”, did you mean even numbers that sum up to six or just any numbers that sum up to 6?

[Corey]

DIsjoint events means that that are independent of each other correct?

[Ezra Halleck]

Danny wrote:
“Disjoint – when two events cannot occur at the same time.”
This is not quite it. Remember, an event is a collection of outcomes. Disjoint means no intersection. So the outcomes in one event are not part of the other event and vice verse.

[Afzal]

a. I still didn’t get it professor.

b. Doubles,P(A)={ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}=6/36=1/6
sum is 6, P(B)={ (1,5),(2,4),(3,3),(4,2),(5,1)}=5/36
P(A and B)={(3,3)}=1/36

c. P(A and B) = P(A) * P(B)
1/36=1/6*5/36
1/36=11/36 which is not true ,so it is not independent.

[Ezra Halleck]

To answer a, the 2 events in question have a common outcome, namely (3,3), so the events are not disjoint.

[soumanusa]

4.
a) A and B are disjoint events.

b) P(A) = 6/36 or 1/6
P(B) = 2/36 or 1/18
P(A & B) = 8/36 or 2/9

c) No, A and B are not independent.
P(A ∩ B) = P(A) * P(B)
0 = 6/36 * 2/36
0 = 1/108
X

[Afzal]

hi soumanusa,
how you get P(B)=2/36?. P(B) is sum of 6. and i think { (1,5),(2,4),(3,3),(4,2),(5,1)} all set comply with the condition. so i think P(B) will be 5/36.

[Ezra Halleck]

Afzal is correct and in fact just 2 posts before your post, soumanusa, he has a correct solution except for a, which I have answered in the post right before yours.