Final answer:
The probability of having a gray-tailed offspring when crossing a homozygous dominant (brown-tailed) squirrel with a homozygous recessive (gray-tailed) squirrel is 0%, as all offspring will be heterozygous and show the dominant brown-tailed phenotype.
Step-by-step explanation:
When considering genetics and inheritance patterns, a brown tail in squirrels being dominant to a gray tail involves simple Mendelian genetics. If a homozygous dominant squirrel (with genotype BB for a brown tail) is crossed with a squirrel with a gray tail (genotype bb, which is homozygous recessive), the offspring will all be heterozygous (Bb) and thus have brown tails since brown is the dominant trait. Consequently, none of the F1 generations will have a gray tail. However, if two heterozygous squirrels (Bb) were crossed, the expected genotype ratio would be 1 BB: 2 Bb: 1 bb, which corresponds to 3 brown-tailed to 1 gray-tailed offspring in terms of the phenotype. Since in the question, one parent is homozygous dominant and the other is homozygous recessive, the probability of an offspring having a gray tail (bb) is 0%.