Thread: Die Event A or B

A six sided die is rolled twice. What is the probability of rolling an even number on the first roll OR a 1 on the second roll? The answer is 7/12 but for the life of me I can't figure this out working backwards/forwards/sideways you name it. Ty

Hi LostInTheSauce, great username....

I call recall a few times where this happened to me!

Anyhow back to the question, let the ordered pair (x,y) be the result of x the first roll and y be the second roll.

Map out all the possibilities.

(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

How many have a even number on the x or a 1 on the y?

Right then 21/36 that works! Just to be certain though there's no way to do this without drawing up a Potential Outcome Square; that is it cannot be solved formulaically? Regardless thanks for the help

Non Mutually Exclusive!

Yeah old news to the world, but brand to this cat. Finally got it in formula fashion with some help.

Subtracting to remove the double count when both A and B are satisfied.

3/6 + 1/6 - 3/36= ALL GLORY TO ZEUS

If events are not mutually exclusive then P(A) or P(B) = P(A) + (PB) - P(A) intersect P(B) - I don't know enough latex to use proper or and intersect symbols, sorry.

Correct, if events A and B are mutually exclusive then