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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?
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
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.
When you say P(A and B), do you mean the union or the intersection?
and=intersection
or=union
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
I see where the problem is now. There are more ways to get a sum of 6, e.g., 1,5.
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?
DIsjoint events means that that are independent of each other correct?
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.
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.
To answer a, the 2 events in question have a common outcome, namely (3,3), so the events are not disjoint.
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
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.
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.
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