https://revisionmaths.com/sites/mathsrevision.net/files/imce/1MA0_1F_que_20160526.pdf Question 5

So what I would do first, would be to write the probability of chosing each counter out of the  bag. For example red is 3/6 => 1/2, green is 2/6 => 1/3, and blue is 1/6.

Then 5a => 1/6 is worse than even but not impossible so it is unlikely.

5b => 1/2 Well there are three red counters and six total meaning prob is 3/6 => 1/2. Hence circle a half.

5c => There are no white counters, so it is impossible!