I need answer Immediately pls!!!!!!

Given:
Total number of students = 27
Students who play basketball = 7
Student who play baseball = 18
Students who play neither sports = 7
To find:
The probability the student chosen at randomly from the class plays both basketball and base ball.
Solution:
Let the following events,
A : Student plays basketball
B : Student plays baseball
U : Union set or all students.
Then according to given information,
[tex]n(U)=27[/tex]
[tex]n(A)=7[/tex]
[tex]n(B)=18[/tex]
[tex]n(A'\cap B')=7[/tex]
We know that,
[tex]n(A\cup B)=n(U)-n(A'\cap B')[/tex]
[tex]n(A\cup B)=27-7[/tex]
[tex]n(A\cup B)=20[/tex]
Now,
[tex]n(A\cup B)=n(A)+n(B)-n(A\cap B)[/tex]
[tex]20=7+18-n(A\cap B)[/tex]
[tex]n(A\cap B)=7+18-20[/tex]
[tex]n(A\cap B)=25-20[/tex]
[tex]n(A\cap B)=5[/tex]
It means, the number of students who play both sports is 5.
The probability the student chosen at randomly from the class plays both basketball and base ball is
[tex]\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}[/tex]
[tex]\text{Probability}=\dfrac{5}{27}[/tex]
Therefore, the required probability is [tex]\dfrac{5}{27}[/tex].