Combinations

Here is one question appeared in CAIE in June 2021(9709/52. Q.6)
I am puzzled by the marking scheme. Please note the question.
Four letters are selected at random from the 8 letters of the word TOMORROW.
Find the provability that the selection contains at least one O and at least one R.
Here the examiner taking 8C4=70 as the total number of combinations . Here I am very frustrated because I have following 15 selections.
OOOR, OORR, OORM, OORW, ORRT , ORRM ,ORRW ,ORTM, ORTW, ORMW OTMW , RRTM , RRMW, RTMW ,RRTW.
IN which 11 sample points contains at least one "O" and at least one "R".
So the probability should be 11/15. But marking scheme answer is 50/70.
My Question is here why we are applying 8C4 in this scenario where letters are repeated. I am also uploading the picture of the Question as well as marking scheme.
Please guide me in this puzzling question.

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Reply 1

mqb2766

You've biased your counting because there is only one way of drawing RRTW, but 3 ways of drawing OORR. Assigning the same probability to each combination isn't correct. You'd be 3 times more likely to pick OORR than RRTW.

You need to assume all the Os and all the Rs are different when counting as each have the same probability, or bias the "probabilities" in your list. Basically the same thing.

(edited 2 years ago)

Reply 2

alevelmath1

Ok! Could you explain why 8C4 in this repeated letters of 8 word.This is my main confusion.

Reply 3

mqb2766

Original post by alevelmath1

Ok! Could you explain why 8C4 in this repeated letters of 8 word.This is my main confusion.

If you treat each letter as distinct 8C4 is the number of groups of 4 (unordered) items you can select from 8, as usual.

But there would then be 3 ways of drawing OORR, for example. So that pattern would occur 3 times in the 8C4 total.