Question

Imagine that two unlinked autosomal genes with simple dominance code in goats for size, where l is large and l is small, and for color, where b is brown and b is white. if a small, white male goat mates with a large, brown female goat of an unknown genotype, what is the probability that they would produce small, white offspring?

Answer 1

1)
T- tall allele;
t- small allele;
B-brown allele
b- white allele

The male has recessive traits for bothe the height gene and the color.
He is small and white and to express these traits it means his genotype  is: ttbb. 
He has both alleles, of each gene, recessive to express his caracteristics.

2) The female genotype is unknown but we know she is tall and brown.
There are 2 possibilities for her genotype:
TTBB or TtBb
In the first case, she would be dominant homozygotic -both alleles, of each gene, are dominant.
The second case, she would carry a recessive allele and a dominant allele. This makes her heterozigotic for both genes.

3)Due to the two possibilies of the female genotype, there will be two different results for the probability of the offspring.
First : ttbb x TTBB- the probability of having small and white offspring would be 0%. That's because the offspring would be: TtBb- expresses the tall and brown traits.

Second: ttbb x TtBb- the probability of having small and white offspring would be 25%. That's because 1/4 of the offspring could have the genotype ttbb.For better understanding i made the punnet square (see the blue part).

Answer 2

1/9 probability of the offspring being small and white.  
Thankfully, I was able to find an identically worded question via Google where the person asking the question didn't perform case folding and as such the correct question starts with:
 "Imagine that two unlinked autosomal genes with simple dominance code in goats for size, where L is large and l is small, and for color, where B is brown and b is white." 
 So we now know that large is dominate and small is recessive. Additionally, we know that brown is dominate and white is recessive. Because of that, the small white male goat is homozygous for the white and small alleles. The large brown female goat, may or may not be homozygous. Since this problem doesn't state the relative frequency of the different alleles, I will assume that they're in the population with a 50/50 probability for both alleles. 
 The small white male goat has a definite genotype of bbww. The large brown female has 9 different possible genotypes. They are bBwW, bBWw, bBWW,BbwW, BbWw, BbWW,BBwW, BBWw, BBWW. Any of the possible combinations for the female are equally possible. Since we're only interested in the probability of a small white offspring, that will be determined solely by the alleles contributed by the female since the male is homozygous in the desired traits. Since there are 9 different possible genotypes for the female and for each genotype, there are 4 possible ways of contributing the alleles, there are potentially 36 different offspring (many of these will be identical in characteristics exhibited). So let's look at them:
 bBwW, bBWw, BbwW, BbWw: Each of these 4 possible genotypes is heterozygous for both traits, so it's possible for an offspring to have both of the desired traits with a 1 in 4 chance for each genotype. So we have 4 possible offspring arising from these genotypes.  
 bBWW, BbWW, BBwW, BBWw, BBWW: All 5 of these genotypes are homozygous for at least one undesired trait. So every possible offspring will be either large, or brown, or both. So there are 0 possible offspring with the desired traits if the mother has any of these 5 genotypes. 
 So of the possible 9*4 = 36 different offspring, 4 of them will have the desired alleles contributed by the mother for a 4/36 = 1/9 probability. 
 Note: In the future when asking a question about genetics. PLEASE PRESERVE THE CASE OF THE QUESTION. Upper case letters represent dominate alleles and lower case represents recessive alleles. Using all lower case renders the question unanswerable. You got lucky in that someone elsewhere asked the exact same question, but was smart enough to preserve the case of the different alleles.