Suppose a single gene controls the color of hamsters: black $(B)$ is dominant and brown $(b)$ is recessive. Hence, a hamster will be black unless its genotype is $b b .$ Two hamsters, each with genotype $B b,$ mate and produce a single offspring. The laws of genetic recombination state that each parent is equally likely to donate either of its two alleles $(B$ or $b),$ so the offspring is equally likely to be any of $B B, B b, b B,$ or $b b$ (the middle two are genetically equivalent).
(a) What is the probability their offspring has black fur?
(b) Given that their offspring has black fur, what is the probability its genotype is $B b ?$
(a) 3/4 $\\$ (b) 2/3
some this problem we've bred to Hampshire's with the capital B lower case be Gino type and were given two questions. One. What's the probability that black fur and two given that they have black for what is the probability that there Gino type is Capital B lower case. Be part, eh? We're gonna use our definition of probability successful outcomes over total outcomes. Now of these forge of these four possible Gino types, this one will be black. This one will be black and this one will be black, which means a probability at a three out of four possible options. Three successes. So we get 3/4 of 0.75 We're part B. We'll just use base there. So this will be probability having black fur given that Gina type times the probability of having that Gino type all over probability. Having black fur. All right. Knowing that let's plug in some numbers the probability of having black for given that your Gina type is Capital B. Lower case be is one. That's how it works. The probability of having that that Gino type is 1/2 because these two are treated as the same. We divide this by the probability of having black fur, which is 3/4 and this ends up being 2/3 0.6 67 ish.