Consider such events:
A - slip with number 3 is chosen;
B - the sum of numbers is 4.
You have to count [tex]Pr(A|B).[/tex]
Use formula for conditional probability:
[tex]Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}.[/tex]
1. The event [tex]A\cap B[/tex] consists in selecting two slips, first is 3 and second should be 1, because the sum is 4. The number of favorable outcomes is exactly 1 and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event [tex]A\cap B[/tex] is
[tex]Pr(A\cap B)=\dfrac{1}{20}.[/tex]
2. The event [tex]B[/tex] consists in selecting two slips with the sum 4. The number of favorable outcomes is exactly 2 (1st slip 3 and 2nd slip 1 or 1st slip 1 and 2nd slip 3) and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event [tex]B[/tex] is
[tex]Pr(B)=\dfrac{2}{20}=\dfrac{1}{10}.[/tex]
3. Then
[tex]Pr(A|B)=\dfrac{\frac{1}{20} }{\frac{1}{10} }=\dfrac{1}{2}.[/tex]
Answer: [tex]\dfrac{1}{2}.[/tex]