Twelve trophies are baseball, eight are football and five are basketball. At random what is the probability of pulling one basketball and tw

There is a total of 25 trophies.

The probability of randomly pulling a basketball trophy is 5 divided by the numbef of trophies remaining.

The probability of randomly pulling a baseball trophy is 12 divided by the number of trophies remaining.

The probability of randomly pulling a second baseball trophy is 11 divided by the number of trophies remaining.

The probability of the trophies randomly chosen in the order  basketball - baseball - baseball is:

     5/25 x 12/24 x 11/23  =  0.0478 (approximately)

The probability of the trophies randomly chosen in the order  baseball - basketball - baseball is:

     12/25 x5/24 x 11/23  =  0.0478 (approximately)

The probability of the trophies randomly chosen in the order  baseball - baseball - basketball is:

     12/25 x 11/24 x 5/23  =  0.0478 (approximately)

Since all are possible and independent of each other, add those probabilities together to get:  0.143

Twelve trophies are baseball,

eight are football and

five are basketball.

At random what is the probability of pulling one basketball and two baseball trophies from boxes

There are 5 ways of pulling out one basketball

and  12C2 = 66 ways of pulling out 2 baseball ones       $${\left({\frac{{\mathtt{12}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{66}}$$ 

Now the total number of ways three trophies can be pulled out is  25C3 =2300

           $${\left({\frac{{\mathtt{25}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{25}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{2\,300}}$$

So    P(1 basketball and 2 baseball) =   (5*66)/2300    approx   0.14

$${\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{66}}}{{\mathtt{2\,300}}}} = {\frac{{\mathtt{33}}}{{\mathtt{230}}}} = {\mathtt{0.143\: \!478\: \!260\: \!869\: \!565\: \!2}}$$

Sorry Geno, My original answer was hidden from your view :)