The tree you drew is not good for all the problems. If the incompetent and forgetful father sends recipes, for each child he has three choices, the real Christmas one $C$, and two others, $A$ and $B$. Each of them has (conditional) probability $\frac{1}{3}$.
List the children in some order, Mike (since he doesn't matter) last. Imagine the e-mails were written in the order Peter, Jane, Mike. So there are $3$ possibilities for Peter. Draw the relevant $3$ branches from the father, and label their ends $A$, $B$, and $C$. From each of these, there is three-way branching to deal with what Jane got. And after that (but we can leave it out), there is three-way branching for Mike.
This more informative tree works fine for the problem you solved: Just look at the paths that lead to $C$ for Peter and for Jane.
For Peter and Jane getting the same recipe, there is of course the branch from the perfect mother. And there are the paths from the father that lead to $A$ for Peter and $A$ for Jane, also the ones that lead to $B$ and $B$, and also the ones that go to $C$ in each case. We get $\frac{1}{2}$ from the mother, and three copies of $\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{3}$ for the father side.
There are many ways to solve the problem. Whether we think of picking the marbles one at a time, or all together, does not alter probabilities, though it will change the way we compute the probabilities.
Imagine the balls are distinct (they all have secret ID numbers). There are $\binom{15}{3}$ equally likely ways to choose $3$ balls from the $15$.
Now we count the number of favourable choices, that is, choices that have $1$ of each colour. There are $\binom{7}{1}\binom{3}{1}\binom{5}{1}$ ways to pick $1$ red, $1$ blue, and $1$ green. Thus our probability is
$$\frac{\binom{7}{1}\binom{3}{1}\binom{5}{1}}{\binom{15}{3}}.$$
Or else we calculate the probability by imagining choices are made one at a time. This complicates things somewhat, since the event "we end up with one of each colour" can happen in various ways.
Let us analyze in detail the probability we get GRB (green then red then blue).
The probability the first ball picked is green is $\frac{5}{15}$ (it is best not to simplify). Given that the first ball was green, the probability the second is red is $\frac{7}{14}$. So the probability the first two picks are green, then red, in that order, is $\frac{5}{15}\cdot\frac{7}{14}$. Given that the first two choices were green and red, in that order, the probability the next is blue is $\frac{3}{13}$. We conclude that the probability of GRB is $\frac{5}{15}\cdot\frac{7}{14}\cdot \frac{3}{13}$.
There is a total of $3$ colour sequences through which we end up with $1$ of each colour. It turns out that each of them has probability $\frac{5}{15}\cdot\frac{7}{14}\cdot \frac{3}{13}$. So our required probability is
$$6\cdot\frac{5}{15}\cdot\frac{7}{14}\cdot \frac{3}{13}.$$
Why were these six probabilities all equal? One way to think about it is as follows. Suppose that, blindfolded, we use a scoop to pull out $3$ balls, and these happen to represent all the colours. Now we reach into the scoop and pull out the balls one at a time. Of course all $3!$ colour orders will be equally likely.
Another more arithmetical way of seeing it is that the denominators will be, in order, $15$, $15$, and $13$, since we are picking from a diminishing number. Each time we consider a partiular colour, the number of choices will be the number of balls of that colour. So the numbers $7$, $3$, and $5$ must each appear exactly once in the numerators.
Best Answer
The following is the tree diagram of the exercise.
The first branching shows the probability of drawing each of the three colors from among the five jelly beans
On the second draw, there are only four beans remaining and one of the three colors will have been reduced in number by one. So whereas the probabilities of the first draw were fractions of $5$, the probabilities of the second draw will be fractions of $4$. The eight results are the probability space of the two draws showing the composition of each event in the space and how the probability of that event is determined. Notice that the sum of all eight probabilities sums to $1$.
You should be able to use this tree to find the answer to all the questions.