How many problems are there?
select count(*) from problems
What products are they associated with?
select * from product as t1 join problems as t2 on t1.product_id = t2.product_id
Now show how many problems exist per product name?
select count(*), t1.product_id from product as t1 join problems as t2 on t1.product_id = t2.product_id group by t1.product_id
Which one has the most problems?
select count(*), t1.product_name from product as t1 join problems as t2 on t1.product_id = t2.product_id group by t1.product_name order by count(*) desc limit 1