A fair die is continuously thrown, what is the probability that a number divisible by two is thrown before a number that is divisible by three is thrown?
The answer is 0.6, but I'm somehow unable to understand why. I assumed that to achieve the above, one cannot throw a 6, and hence, as the numbers that are to be considered here would be {2,3,4,6}, and what I got was 2/4, which isn't correct. Would be grateful if someone could give me pointers/tell me what the correct way to calculate this is! Thank you!