What is the probability of landing on a 4 and then landing on a 7

Let's say:
A: landing on a 4
B: landing on a 7
We have that the probability of each event is:
P(A) = 1/4
P(B) = 1/4
because it would be 1 out of four events.
Since they are independent events ( this means that it doesn't matter if we land on a 4 before, it will not affect the next result), the probability of landing on a 7 and landing on a 4 is given by
[tex]\begin{gathered} P\mleft(AandB\mright)=P(A)\cdot P(AB) \\ P\mleft(AandB\mright)=\frac{1}{4}\cdot\frac{1}{4} \\ P(AandB)=\frac{1}{16} \end{gathered}[/tex]When the events are independent they both happening has a probability given by the product of each probability
Answer: P(AandB) = 1/16