Solution

 

a. An offspring will have smooth yellow peas if there is at least one $S$ and at least one $Y$ in gene pairs for first and second trait, respectively. There are 9 such gene pairs. Since, each pairing is equally likely, probability of smooth yellow peas is $\dfrac{9}{16}.$

b. An offspring will have smooth green peas if it has at least one $$S for first trait and $$yy gene pair for second trait. There are 3 such gene pairs. Since, each pairing is equally likely, probability of smooth green peas is $$3/16

sc. An offspring will have wrinkled yellow peas if it has $ss$ gene pair for first trait and at least one $Y$ in gene pair for second trait. There are 3 such gene pairs. Since, each pairing is equally likely, probability of wrinkled green peas is $\dfrac{3}{16}.$

d. An offspring will have smooth green peas if it has $$ss gene pair for first trait and $$yy gene pair for second trait. There is 1 such gene pair. Since, each pairing is equally likely, probability of wrinkled green peas is1/16.

e. Let $L$, $M$ be the event that offspring has smooth and yellow peas, respectively. $L$ and $M$ are independent.

Let $B$ and $C$ be the event that offspring carries one $s$ allele and offspring carries one $s$ allele and one $y$ allele, respectively.

$M$ is independent of $B.$

From solution to previous part, we know that $P(L \cap M)= \dfrac{9}{16}.$

Since, all pairings are equally likely,

$$$$P(B)= \dfrac{1}{2}, ~~P(C)= \dfrac{1}{4}, ~~P(M)= \dfrac{3}{4}

Since, $Y$ and $S$ are dominant traits, we would have

$$$$P(L|B)=1, ~~P(L \cap M|C)=1.

$Answer$