The color of a person’s eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact, we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith’s sister has blue eyes.

(a) What is the probability that Smith possesses a blue eyed gene?

(b) Suppose that Smith’s wife has blue eyes. What is the probability that their first child will have blue eyes?

(c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?

$B$- gene for brown eyes (dominant)

$b$- gene for blue eyes (not dominant)

a person has two genes for eye-color -$(m,f)$

$m$form mother (any of her two genes with equal probability)

$f$from father (any of his two genes with equal probability)

Smith, brown eyes$\to (B,B),(b,B),(B,b)$

Smith's sister, blue eyes$\to (b,b)$

Smith's mother, brown eyes$\to (B,B),(b,B),(B,b)$

Smith's father, brown eyes $\to (B,B),(b,B),(B,b)$

For every individual noted, each of the gene combinations is equally probable.

We consider that for the parents, every combination in children's genes is one precise preference from the parent's genes.

a) $P(Smith=(b,B)$or $Smith=(B,b))=?$

According to the text above, all that is left is to add the probabilities of two mutually exclusive possibilities:

localid="1647096814612" $P(\text{Smith}=(B,b\left)\right)=P(\text{Smith}=(b,B\left)\right)=P(\text{Smith}=(B,B\left)\right)=\frac{1}{3}$

localid="1647096823182" $P(\text{Smith}=(b,B)\text{or Smith}=(B,b\left)\right)=\frac{2}{3}$

Smith, brown eyes$\to (B,B),(b,B),(B,b)$

Smith's sister, blue eyes$\to (b,b)$

Smith's mother, brown eyes$\to (B,B),(b,B),(B,b)$

Smith's father, brown eyes$\to (B,B),(b,B),(B,b)$

b) Smith's wife$=(b,b)-$blue eyes, $P($Smith ${}^{\text{'}}s$child $\left.=(b,b)\right)=?$

This event can be broken down into cases by whether Smith has the blue eyes gene (call this, the event $A$, from $a$), localid="1646405230387" $P\left(A\right)=\frac{2}{3}$)

localid="1647096835267" $P\left({\text{Smith}}^{\text{'}}\text{s child}=(b,b)\right)=P\left({\text{Smith}}^{\text{'}}s\text{child}=(b,b)\mid A\right)P\left(A\right)+$

localid="1647096841518" $P\left({\text{Smith's}}^{\text{'}}\text{child}=(b,b)\mid A^c\right)P\left(A^c\right)$

The mother surely passes on the blue eyes gene, and the father with probability localid="1647096850623" $\frac{1}{2}$if he has a blue eyes gene, and localid="1647096858296" $0$if he does not. The child only has blue eyes if it inherited two blue eyes genes, therefore:

localid="1647096864349" $P\left(Smit{h}^{\text{'}}s\text{child}=(b,b)\right)=\frac{1}{2}·\frac{2}{3}+0·\frac{1}{3}=\frac{1}{3}$

Smith, brown eyes$\to (B,B),(b,B),(B,b)$

Smith's sister, blue eyes$\to (b,b)$

Smith's mother, brown eyes$\to (B,B),(b,B),(B,b)$

Smith's father, brown eyes$\to (B,B),(b,B),(B,b)$

Consider "A" is event that Smith have blue eyed gene so

From part a $P\left(A\right)=\frac{2}{3}$

As per the result from part b, The probability of first child with blue eye is $\frac{1}{3}$

Consider F is event that smith first child have brown eye. So,

$P\left(F\right)=1-\frac{1}{3}=\frac{2}{3}$

Hence, $P(\mathrm{B}\mid F)=\frac{P(\mathrm{B}\cap F)}{P\left(F\right)}=\frac{P(F\mid \mathrm{B})P\left(\mathrm{B}\right)}{P\left(F\right)}=\frac{1\left({\displaystyle \frac{1}{3}}\right)}{{\displaystyle \frac{2}{3}}}=0.5$

$P(A\mid F)=1-P(B\mid F)=1-0.5=0.5$

Consider Gis event that second child have brown eye.

Therefore,

$P(G\mid B,F)=1$

$\mathrm{P}(\mathrm{G}\mid \mathrm{A},\mathrm{F})=\frac{1}{2}=0.5$

We need to find $\mathrm{P}(\mathrm{G}\mid \mathrm{F})$

Now,

$P(G\mid F)=P(G\cap B\mid F)+P(G\cap A\mid F)$

$=P(G\mid F,B)\times P(B\mid F)+P(G\mid F,\mathrm{A})\times P(\mathrm{A}\mid F)$

$=1\times 0.5+0.5\times 0.5=0.75$