Hi Ginger,

5 of the 9 letters of the word MINCEMEAT are selected.

Find the number of possible selections which contain exactly 1 M and 1 E.

I just mean that if 5 letters are selected at random then the prob of drawing MECAT is higher then the prob of drawing INCAT (becasue there are 2Ms and 2Es to chose from)

So saying there are 10 distinct outcomes is correct but each one does not have the same probability of occuring as other possible outcomes.

eg

What is the probability of chosing a selection of 5 letters that contain exactly one M and one E

We have determined that there are 10 such unique combinations

How many combinations (not unique) are there altogether. (think in terms of M1, M2, E1, E2)

9C5 = 126

Now how many of these have exactly one M and one E

M1 OR M2, E1 or E2, plus 3 more.

2*2*5C3 = 40

So I think that if 5 letters are selected at random then the prob of having exactly one M and one E is 40/126 = 20/63