Probability of equal # of blue and red fish chosen when drawing without replacement

Question

The problem is such

One diagram of the possible situations is as follows:

M M F F , $P(M=2,F=2)$

Need to find , $ P(M=2,F=2)$, So:

-# of ways to pick four fish from the pond = $C_2^{10}C_2^{8}=45*28=1260$

-# of ways to pick 2 male and female fish from the pond = $(10*9)+(8*7)=90+56$

So the probability is $\frac{146}{1260}=0.11589$, which seems a bit too small; shouldn't it be closer to 1 as what other cases would make up for the remaining 1114(=1260-146) cases. I am guessing that I went wrong with finding the possible ways as 146. And I am not sure if I need to consider if there was only 1 male and female fish and 0 male and female fish. Cause it doesn't make sense that when you draw 4 fish u get none or only 2 fish.

Answer 1

You have to consider whole items.
Now Total = ¹⁸C₄ = 3060
Like you said, ways to pick 2 male and female fish from the pond = ¹⁰C₂ * ⁸C₂ = 1260
So the probability is ¹²⁶⁰ ⁄ ³⁰⁶⁰=0.4117