Finding probability

The probability that $3$ occurs before $5$ , with $3$ occurring on the $n$ th trial, is $\big(\tfrac58\big)^{n-1} \times \tfrac18$ . Thus the probability that $3$ occurs before $5$ is $\sum_{n=1}^\infty \big(\tfrac58\big)^{n-1} \times \tfrac18 \; = \; \boxed{\tfrac13}$

Let

event A={"3" occurs in the first experiment}, event B={"5" occurs in the first experiment}

event C={neither "3" nor "5" occurs in the first experiment}, event D={"3" occurs before "5" in a row of experiments}

It is simple to know:

$P(A)=\frac{2}{4\times 4}=\frac{1}{8}, P(B)=\frac{4}{4\times 4}=\frac{1}{4}, P(C)=1-P(A)-P(B)=\frac{5}{8}$

since:

$A\bigcup B\bigcup C=S, A\bigcap B=AB=\varnothing, A\bigcap C=\varnothing, B\bigcap C=\varnothing$

we get:

$(A\bigcup B\bigcup C)\bigcap D=AD\bigcup BD\bigcup CD=S\bigcap D=D,\\ AD\bigcap BD=\varnothing, AD\bigcap CD=\varnothing, BD\bigcap CD=\varnothing$

finally,

$P(D)=P(AD)+P(BD)+P(CD)=P(A)*P(D|A)+P(B)*P(D|B)+P(C)*P(D|C)$

Here S represents a certain event.

In fact, it's not complicated to get this formula just using your intuition.

Also,

$P(D|A)=1, P(D|B)=0, P(D|C)=P(D)$

Note that different trails are independent, so the first time you didn't obtain a "3" or "5"(that is, event C happens), the probably that D happens remains the same. $P(D)=\frac{1}{8}\times 1+\frac{1}{4}\times 0+\frac{5}{8}\times P(D)\rightarrow P(D)=\frac{1}{3}\approx0.333$