Answer:
0.5781
Step-by-step explanation:
Let 'S' be the dominant allele for solid coat color and 's' be the recessive allele for spotted coat color.
By crossing a homozygous solid-colored dog (SS) with a spotted dog (ss), all of the F1 will present the Ss genotype.
When crossing two F1 dogs, the sample space for F2 is:
F2={Ss,sS,ss,SS}
For each puppy, there is a 1 in 4 chance of having a spotted coat.
The probability that at least one of the three F2 puppies have a spotted coat is 100% minus the probability that none have a spotted coat:
[tex]P(ss>0) =1- P(ss=0)\\P(ss>0) =1- (1-\frac{1}{4} )^3\\P(ss>0) = 0.5781[/tex]
The probability that at least one of the three F2 puppies will have a spotted coat is 0.5781.
