ANSWER
[tex]P(A|B)=\frac{1}{3}\approx0.333[/tex]
EXPLANATION
We want to find the probability P(A|B).
This is a conditional probability and its formula is:
[tex]P(A|B)=\frac{P(A\cap B)}{P(B)}[/tex]
where P(A n B) is the probability that the student plays basketball and soccer and P(B) is the probability that the student plays soccer.
We have that:
[tex]P(A\cap B)=\frac{1}{10}[/tex]
and
[tex]P(B)=\frac{3}{10}[/tex]
Therefore, we have that:
[tex]\begin{gathered} P(A|B)=\frac{1}{10}\div\frac{3}{10} \\ P(A|B)=\frac{1}{10}\cdot\frac{10}{3} \\ P(A|B)=\frac{1}{3}\approx0.333 \end{gathered}[/tex]
That is the answer.
