Answer: 3
Step-by-step explanation:
Let E be the event of that student pierces ear and N be the event of that student pierces nose.
Given: [tex]n(E\cup N=50)[/tex]
[tex]n(E)=46\\\\n(N)=7[/tex]
For any two event A and B, we have
[tex]n(A\cup B)=n(A)+n(B)-n(A\cap B)[/tex]
Similarly , [tex]n(E\cup N)=n(E)+n(N)-n(E\cap N)[/tex]
[tex]50=46+7-n(E\cap N)\\\\\Rightarrow\ n(E\cap N)=53-50=3[/tex]
Hence, 3 students have piercings both on their ears and their noses.