Q:

Conditional Probability - Finding $P(A)$

During sports week at a local school, the school nurse recorded information about injuries.
Let A be the event “a student is injured playing tennis”.
Let B be the event “a student is injured playing netball”.
From the information the school nurse has recorded, it can be deduced that:
    $P(A ∪ B) = 0.35$ and $P(A ∪ B') = 0.90.$
             Determine the proportion of students who were injured playing tennis during sports
    week.

My working: I know that $P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$
So from the information given we can get
$P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.35$
and $P(A ∪ B') = P(A) + P(B') - P(A ∩ B') = 0.90 \Leftrightarrow P(A) + 0.1 -P(A ∩ B') = P(B)$
and we are trying to find $P(A)$ so we can form some simultaneous equation as such:
$P(B) = 0.35-P(A)-P(A ∩ B)$
and $P(B)=  P(A) + 0.1 -P(A ∩ B')$
Which implies: $$0.35-P(A)-P(A ∩ B) =P(A) + 0.1 -P(A ∩ B')$$
$$ \Leftrightarrow P(A ∩ B')-P(A ∩ B)=2P(A)-0.25 $$
Now I am stuck as there are two unknowns, I would also like to learn other methods of solving this question - I think you can do this question with venn diagrams but I'm unsure how to set them up..

A:

Define:  $A_B=A\cap B$ and $A_{B'}=A\cap B'$.  These are disjoint and $A=A_B \cup A_{B'}$ so $P(A)=P(A_B)+ P(A_{B'})$.
We are told that $$.35=P(A\cup B)=P(A_{B'}\cup B)=P(A_{B'})+P(B)$$
$$.9=P(A\cup B')=P(A_B\cup B')=P(A_B)+ P(B')=P(A_B)+1-P(B)$$
add to obtain:  $$1.25=P(A_{B'})+P(A_B)+1\implies P(A)=.25$$

