In labrador retrievers, a common breed of dog, black coat is dominant to chocolate, normal vision is dominant to progressive retinal atrophy (pra), and normal hip joint is dominant to hip dysplasia. all these genes assort independently. two dogs that are heterozygous for alleles of all three genes are crossed. using rules of probability (not a punnett square), what is the chance that the first pup born to these dogs will be chocolate, have normal vision, and have normal hip joints?

The data given regarding the alleles are as follows
black(B)-dominant , chocolate(b) - recessive
normal vision - dominant, progressive retinal atrophy - recessive 
normal hip joint - dominant, hip dysplasia - recessive 

lets first consider one trait 
parents are heterozygous 
when crossed            Bb         x          Bb
offspring               BB      Bb     Bb     bb
                             black  chocolate
the probability of dogs being black - 3/4
                           dogs being chocolate - 1/4
likewise for the other two traits 
probability of normal vision - 3/4 and retinal atrophy - 1/4
probability of normal hip joint - 3/4 and hip dysplasia - 1/4

we have to calculate the probability of 1st pup being born is
chocolate - 1/4 
normal vision - 3/4
normal hip - 3/4
since all these traits should be acquired the probability of having the above three traits is calculated by getting the product of individual probabilities 
therefore answer is 1/4 x 3/4 x 3/4
  probability              = 9/64

The answer is going to be a because ik that i’m right