Two Events A & B: Is r+s >1 Possible? - Brendan's Query

[brendan]

I have read a question which has

Two events A and B such that p(A) = r and p(B) = s with r,s >0 and
r + s > 1

My querry is about r + s > 1 is this possible ? I thought that it had to be < or = to 1


regards
Brendan

 

[CompuChip]

Suppose you have a die, and you throw it once. Let A be "the outcome is 1, 2, 3, 4 or 5" and B "the outcome is 3, 4, 5 or 6".

Note however, that though p(A) + p(B) > 1, you always have p(A + B) <= 1 (with p(A + B) being the probability of A or B happening).

 

[HallsofIvy]

I have read a question which has

Two events A and B such that p(A) = r and p(B) = s with r,s >0 and
r + s > 1

My querry is about r + s > 1 is this possible ? I thought that it had to be < or = to 1


regards
Brendan

If A and B are "mutually exclusive" events, then P(A+ B)= P(A)+ P(B). Since P(A+B) cannot be larger than 1, neither can P(A)+ P(B).

But if A and B are NOT "mutually exclusive" that is not true. Suppose you roll a single die. Let A be "the number on the die is larger than 2" and B is "the number on the die is even"
Since there are 4 numbers on the die larger than 2, P(A)= 4/5= 2/3. Since there are 3 even numbers on the die, P(B)= 3/6= 1/2. P(A)+ P(B)= 2/3+ 1/2= 7/6> 1.

Of course, that is NOT P(A+B). The numbers on the die that are "either larger than 2 or even" are 2, 3, 4, 5, 6 so P(A+ B)= 5/6.

"Larger than 3 or even" are not mutually exclusive. 4 and 6 are both "larger than 3" and even. P(A and B)= 2/6= 1/3 so P(A+ B)= P(A)+ P(B)- P(A and B)= 2/3+ 1/2- 1/3= 7/6- 2/6= 5/6.

 

[brendan]

Thanks very much guys.
I see that it makes sense that they could both equal > 1 if both are not mutually exclusive.

Now I just have to prove it !

 

[John Creighto]

Note however, that though p(A) + p(B) > 1, you always have p(A + B) <= 1 (with p(A + B) being the probability of A or B happening).

Don't you mean:

p(A + B) <= P(a)+P(b)

which is a triangle inequality.

 

[CompuChip]

Yes, that is also true. Actually I think it is something like
p(A \cup B) = p(A) + p(B) - p(A \cap B)
which shows both the triangle inequality and the statement about mutually exclusive events (where A \cap B is empty).

However, the point I wanted to stress is that the probability of A or B is always less than or equal to one, as you'd expect, contrary to the probability of A plus the probability of B; therefore p(A) + p(B) is not the same as p(A + B).