# Problem regarding genes and conditional probability.

I've been trying to find a satisfactory answer to this problem, but I can't seem to convince myself that my answer is right. I'm having problems with the last item, but I'm transcribing the whole problem here since my error may came from an earlier item.

Eye color is determined by 2 genes. If they both codify for blue eyes, then the person has blue eyes, and if not then the person has brown eyes. We suppose eyes can be either blue or brown and a baby receives one gene from it's mother and one from it's father and each gene from a parent has the same probability of being transmitted to the child. Now let's suppose Brad's parents both have brown eyes but his sister has blue eyes. Let's call the gene that codifies for blue eyes A and the one that codifies for brown ones B.

1. What's the probability that Brad has the gene that codifies for blue eyes?

Since we know that both his parents have both genes, then there's 4 options and they are equiprobable. $P(\text{Brad has the gene A}) = P(Brad \in \{AB,BA,AA\})= \frac{3}{4}$

2. Let's suppose that Brad has brown eyes. Does the probability that he has gen A changes?

It does, since $P(\text{Brad has the gene A knowing he has gene B}) = P(Brad \in \{AB,BA\})= \frac{2}{3}$

1. Now Brad (who we know has brown eyes) has 2 sons with Anna, who has blue eyes. Show that the probability of each one having blue eyes is the same and find the value.

Using total probability and calling $S_1$ and $S_2$ his 2 sons, we know that $$P(S_1 = AA) = P(S_1 = AA | Brad \in \{AB, BA\} ) P(Brad \in \{AB,BA\}) + P(S_1 = AA | Brad = BB ) P(Brad = BB)$$

$$P(S_1 = AA) = \frac{1}{2} \times \frac{2}{3} + 0 \times \frac{1}{3} = \frac{1}{3}$$

This doesn't change for each son.

1. Calculate the probability that one has brown eyes given that the other does. Are the events $\{S_1 \text{has brown eyes}\}$ and $\{S_2 \text{has brown eyes}\}$ independent?

It does specify which one, so let's say I want to find $$P(S_2 \in \{AB, BA, BB \} | S_1 \in \{ AB, BA, BB \})$$

Since their mother has blue eyes, we know that they can't have BB, so we're actually looking for

$$P(S_2 \in \{AB, BA\} | S_1 \in \{ AB, BA\})$$

By Bayes Theorem

$$P(S_2 \in \{AB, BA\} | S_1 \in \{ AB, BA\}) = \frac{P(S_1 \in \{AB, BA\} | S_2 \in \{ AB, BA\}) P(S_2 \in \{AB, BA\})}{P(S_1 \in \{AB, BA\})}$$

Here's where I find a problem and don't know how to continue.

I've tried the following, but I realize it's not right.

Since we're assuming $S_1 \in \{ AB, BA\}$ already then

$$P(S_1 \in \{AB, BA\}) = 1$$

So we get,

$$P(S_2 \in \{AB, BA\} | S_1 \in \{ AB, BA\}) = \\ P(S_1 \in \{AB, BA\} | S_2 \in \{ AB, BA\}) P(S_2 \in \{AB, BA\})$$

And that's as far as I can get. I imagine there's something about conditional probability that I'm missing, but I've reviewed my notes a few times and haven't been able to find it.

Any help would be greatly appreciated.

• $P(S_1\in\{AB,BA\})$ is your prior probability, meaning that it represents the overall probability that $S_1$ has brown eyes. So even though we're assuming that $S_1$ does have brown eyes, the probability itself is not $1$. – Brent May 12 '15 at 17:21
• Right, of course. What I'm having trouble working with is $P(S_1 \in \{AB,BA\}|S_2 \in \{AB,BA\})$, since I can't make any of the prior stuff fit in there. Thanks for the comment. – John Williams May 12 '15 at 17:24

Well, after a few days of thinking about it, I post and then manage to figure it out. I guess I'll leave the answer here and maybe someone else can use it.

First, let's call $M = \{AB, BA \}$. So, we're looking for $P(S_2 \in M | S_1 \in M)$ since we already know that they have a blue gene.

Now, by total probability

$$P(S_2 \in M | S_1 \in M) = P(S_2 \in M | Brad\in M ; S_1 \in M)P( Brad \in M | S_1 \in M) + P(S_2 \in M | Brad = BB ; S_1 \in M)P(Brad = BB | S_1 \in M)$$

Its clear that $P(S_2 \in M | Brad = BB ; S_1 \in M) = 1$ and $P(S_2 \in M | Brad\in M ; S_1 \in M) = \frac{1}{2}$, so we only need to find the other two terms.

Remember we already know that $P(S_1 \in M) = 1 - P(S_1 = AA) = \frac{2}{3}$

$$P( Brad \in M | S_1 \in M) = \frac{P(S_1 \in M | Brad \in M)P(Brad \in M)}{P(S_1 \in M)} = \frac{\frac{1}{2} . \frac{2}{3}}{\frac{2}{3}} = \frac{1}{2}$$

$$P( Brad = BB | S_1 \in M) = \frac{P(S_1 \in M | Brad = BB)P(Brad = BB)}{P(S_1 \in M)} = \frac{1 . \frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$$

Going back to the beginning: $$P(S_2 \in M | S_1 \in M) = \frac{1}{2} . \frac{1}{2} + 1 . \frac{1}{2} = \frac{3}{4}$$

Since $P(S_2 \in M) = \frac{2}{3}$ the events are not independent.