Statistics and the addition rule. i dont get why 2/3*2/3 is not 4/6... plus if i am to use the addition rule, it will be 2/3+2/3 -(The correct answer)  what will that be then?

The question was.P((3 or 5 on the first toss of a die) or (3 or 5 on the second toss of a die)) and i seem to be getting stuck.
$$
\frac 2 3 \times \frac 2 3 = \frac {2\times2}{3\times3} = \frac 4 9.
$$

The probability of $3$ or $5$ on the first trial is $1/3.$

The probability that that happens on the first toss or the second is not $\frac 1 3 + \frac 1 3$ because those two events are not mutually exclusive: they could both happen.

One way to find the probability that one of these or the other (or both) occurs is to find the probability that they both fail to occur. The probability that the first one fails is $2/3$ and the probability that the second one fails is $2/3.$ The probability that the first one fails and the second one fails is $\frac 2 3 \times \frac 2 3 = \frac 4 9.$ Therefore the probability that you don't get two failures, i.e. that you do get at least one success, is $1- \frac 4 9 = \frac 5 9.$
$$
\frac 2 3 \times \frac 2 3 = \frac {2\times2}{3\times3} = \frac 4 9.
$$

The probability of $3$ or $5$ on the first trial is $1/3.$

The probability that that happens on the first toss or the second is not $\frac 1 3 + \frac 1 3$ because those two events are not mutually exclusive: they could both happen.

One way to find the probability that one of these or the other (or both) occurs is to find the probability that they both fail to occur. The probability that the first one fails is $2/3$ and the probability that the second one fails is $2/3.$ The probability that the first one fails and the second one fails is $\frac 2 3 \times \frac 2 3 = \frac 4 9.$ Therefore the probability that you don't get two failures, i.e. that you do get at least one success, is $1- \frac 4 9 = \frac 5 9.$