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

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- Mar 16th 2011, 07:51 PMLostInTheSauceDie 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

- Mar 16th 2011, 07:58 PMpickslides
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? - Mar 16th 2011, 08:08 PMLostInTheSauce
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

- Mar 17th 2011, 02:20 PMLostInTheSauceEureka!
**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 - Mar 18th 2011, 10:08 AMangypangy
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.

- Mar 18th 2011, 01:02 PMpickslides
Correct, if events A and B are mutually exclusive then $\displaystyle \displaystyle P(A \cap B ) = 0$