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I am stumped...please help...I'm not sure what they are talking about with P (A and B) = 0

If P(A) = 2/3, P(B) = 1/6, and P(A and B) = 0, what can you say about P (A or B)?

2 answers

Have you come across the formula
P(A or B) = P(A) + P(B) - P(A and B)
so we have
P(A or B) = 2/3 + 1/6 - 0 = 5/6

let me illustrate with an example:
Sam has 12 identical boxes in front of him. Inside 8 of the boxes are placed even numbers, inside 2 of them are placed odd numbers and inside the last two boxes are placed red marbles.

One box is picked at random.

Let A be: picking an even number.
Let B be: picking an odd number.

So P(A) = 8/12 = 2/3
P(B) = 2/12 = 1/6
P(picking an even AND an odd) = P(A and B) = 0 (NO WAY can that happen)

But picking an even OR an odd
= P(A or B) = 10/12 = 5/6
(there are 10 numbers in the boxes, either even or odd)

Now back to
P(A or B) = P(A) + P(B) - P(A and B)
= 2/3 + 1/6 = 5/6
Thank you, that really helped me!