A die is loaded so that the numbers 2 through 6 are equally likely to appear but that 1 is 3 times as likely as any other number to appear. What is the probability of getting an even number?
This is where I become lost - I'm not sure how to get the outcome.
The Attempt at a Solution
P(2) + P(4) + P(6) = 1/8 + 1/8 + 1/8 = I think the answer is 3/8 but can't tell you how I got that. I don't understand the P(1) + P(2) + P(3) thing. Could someone help me to understand this?