Probability of 'at least' event In a roulette of letters A and B, the probability of choosing A is 0.92, B is 0.33, and both is 0.29. What is the probability of choosing at least 1 of the letters?

P(at least 1) = P(A) + P(B) - P(A and B) = 0.92 + 0.33 - 0.29 = 0.96

I am confused if this is the right way to solve P (at least 1), that is, similar to P(A U B). I have consulted other references but they do not give examples with given probabilities, particularly P(both).

The $P(\text{at least }1)$ is exactly $P(A \cup B)$. To calculate $P(A \cup B)$, when you add up $P(A)$ and $P(B)$, the part they share, i.e, $P(A \text{ and }B)$, has been added for two times so we subtract $P(A \text{ and }B)$ from the sum.
