Joining 2 tables with Group By - mysql

Question

poll_opts table for storing options and poll_voted for storing vote result, pid stands for poll id(unique) and oid stands for option id(unique for individual poll only)
poll_voted [Primary Key: pid.oid.emp]
+-----+-----+-------+
| pid | oid | emp |
+-----+-----+-------+
poll_opts [Primary Key: pid.oid]
+-----+-----+---------+
| pid | oid | opt |
+-----+-----+---------+
pid & oid type: int , opt type: text

Answer 1

If you need the "not existent" results as well you need a left outer join preserves all results from poll_opts even if no match in poll_votes is found.
MySql 5.7 Join Syntax
Query:
select opt, count(vo.oid)
from poll_opts po
left outer join poll_voted vo on vo.oid = po.oid and po.pid=vo.pid
where po.pid = 3 -- 3
group by opt
Output:
opt count(vo.oid)
Chrome 0
Firefox 0
IE 0
MS Edge 0
Opera 1
Testdata:
CREATE TABLE poll_voted (`pid` int, `oid` int, `emp` int);
INSERT INTO poll_voted (`pid`, `oid`, `emp`)
VALUES
(1, 0, 1989),
(1, 2, 1989),
(1, 4, 1989),
(1, 6, 1989),
(3, 2, 1989)
;
CREATE TABLE poll_opts (`pid` int, `oid` int, `opt` varchar(15));
INSERT INTO poll_opts (`pid`, `oid`, `opt`)
VALUES
(1, 0, 'WinXP'),
(1, 2, 'wIN7'),
(1, 4, 'wIN 10'),
(1, 6, 'Ubuntu'),
(3, 0, 'IE'),
(3, 1, 'MS Edge'),
(3, 2, 'Opera'),
(3, 3, 'Chrome'),
(3, 4, 'Firefox')
;