Question 923983
There are
{{{9*8*7*6/(1*2*3*4)=126}}} possible groups of 4 mice.
Some of those include Mike Mouse.
How many groups of 4 mice including Mike Mouse are there?
As many as groups of 3 mice made from the other eight mice.
That is {{{8*7*6/(1*2*3)=56}}}
Of the total of {{{126}}} possible groups of 4 mice,
there are {{{56}}} such groups including Mike Mouse.
The probability that Mike mouse will be included is
{{{56/126=highlight(4/9="0.4444..."="44%(rounded)")}}}
 
A different way:
As you select the first mouse,
the probability of not selecting Mike Mouse is {{{8/9}}} .
After that, the probability of not picking Mike as your second Mouse is {{{7/8}}} .
After that, the probability of not picking Mike as your third Mouse is {{{6/7}}} .
And, if you have not picked Mike Mouse as one of the first 3 mice you pick,
the probability of not picking Mike as your fourth Mouse is {{{5/6}}} .
All in all, the probability of not picking Mike Mouse as one of you four control mice is
{{{(8/9)*(7/8)*(6/7)*(5/6)=8*7*6*5/(9*8*7*6)=cross(8)*cross(7)*cross(6)*5/(9*cross(8)*cross(7)*cross(6))=5/9}}} .
So the probability of that not happening, and having Mike in your control group is
{{{1-5/9=highlight(4/9)}}} .