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In a case where black dominates over brown, a black rat can have one of the following two genotypes BB (pure black with two black genes) or Bb (hybrid with one black and one brown gene). Brown rats will have the genotype bb (two brown genes). The fact that a black rat with two black parents has a brown sibling suggests that the parents are hybrid. If any parent had been pure black (BB), they wouldn't be able to produce a brown offspring since they would only pass on a black gene (B) to any of their offsprings. Therefore, the parents must have the genotypes Bb.

To calculate the probability of the rat being a pure black rat, we need to consider all the possible genotypes it can have and find the probability of having a BB genotype. When a hybrid rat (Bb) mates with another hybrid (Bb), the offspring can have 4 different combinations of genotypes: BB, Bb, bB, and bb. Let's find the probability of the offspring being BB given that it’s a black rat: P(BB | Black Rat) = \( \frac{P(BB \cap Black Rat)}{P(Black Rat)} \) Since all BB rats are black, P(BB | Black Rat) = 1. Now, we need to find the probability of the offspring being a black rat: P(Black Rat) = P(BB or Bb or bB) = P(BB) + P(Bb) + P(bB) Since the parents are hybrids (Bb), the probability of each genotype outcome is: P(BB) = \( \frac{1}{4} \) P(Bb) = \( \frac{1}{4} \) P(bB) = \( \frac{1}{4} \) P(bb) = \( \frac{1}{4} \) Then, P(Black Rat) = \( \frac{1}{4} \) + \( \frac{1}{4} \) + \( \frac{1}{4} \) = \( \frac{3}{4} \) Now, we can find the probability of the offspring being a pure black rat (BB): P(BB | Black Rat) = \( \frac{P(BB)}{P(Black Rat)} \) = \( \frac{\frac{1}{4}}{\frac{3}{4}} \) = \( \frac{1}{3} \) Thus, the probability that the rat is a pure black rat (BB) is \( \frac{1}{3} \).

Now that we know that when the black rat is mated with a brown rat (bb), all five offspring are black, we need to find the probability that the rat is a pure black rat considering this additional information. Let's label the events: Event A: The rat is a pure black rat (BB) Event B: All five offspring are black Now, we need to find P(A | B), i.e., the probability that the rat is a pure black rat given that all five offspring are black. P(A | B) = \( \frac{P(A \cap B)}{P(B)} \) From step 2, we know that: P(A) = \( \frac{1}{3} \) P(B | A) represents the probability that all five offspring are black when the black rat is a pure black. Since a pure black rat can only pass the black gene to its offspring, P(B | A) = 1. P(B | A') represents the probability that all five offspring are black when the black rat is not pure black (Bb). A hybrid rat (Bb) mating with a brown rat (bb) can produce black (Bb) or brown (bb) offspring with equal probability. Thus, P(B | A') = \( (0.5)^5 \) = \( \frac{1}{32} \) P(A') = 1 - P(A) = \( \frac{2}{3} \) Now, we can find P(A ∩ B) and P(B): P(A ∩ B) = P(A) * P(B | A) = \( \frac{1}{3} \) * 1 = \( \frac{1}{3} \) P(A' ∩ B) = P(A') * P(B | A') = \( \frac{2}{3} \) * \( \frac{1}{32} \) = \( \frac{1}{48} \) P(B) = P(A ∩ B) + P(A' ∩ B) = \( \frac{1}{3} \) + \( \frac{1}{48} \) = \( \frac{17}{48} \) Finally, we can find P(A | B): P(A | B) = \( \frac{P(A \cap B)}{P(B)} \) = \( \frac{\frac{1}{3}}{\frac{17}{48}} \) = \( \frac{16}{17} \) Thus, the updated probability that the rat is a pure black rat given that all five offspring are black is \( \frac{16}{17} \).