If you have $9$ apples, you can choose $0, 1, 2, 3, ... 9$ of them. So there are $10$ ways to choose. Similarly for oranges, you have up to $7$ ways to choose. By the fundamental counting principle, you have $10\times7 = 70$ possible ways of choosing fruits (or no fruit). There is only one way to choose no fruit so we subtract $1$.
If you've used generating functions, you could also see it but it would also be a hassle to expand.
Also, have you ever encountered a problem such as "How many divisors does x have? (x non-prime)" It is similar to that where you add $+1$ to each power of the prime factorization when multiplying them.
If each apple and each orange were distinguishable, there would again be only one way to choosing nothing.
If the $9$ apples were distinguishable and the $6$ oranges were distinguishable, I believe this is the same as considering $9 + 6 = 15$ different objects.
How many ways are there to choose from $15$ different objects? There are $2^{15}$ such ways (including not choosing anything). Then $2^{15}-1$ is gives you the number of ways in which you have to choose at least one fruit. This is analogous to a power set.
Easy way:
Since it is given that both pieces of fruit are yellow, pretend that the green fruit do not exist.
Probability is
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}$$
where $D = \binom{11}{2}$.
Then $N = \binom{5}{1} \times \binom{6}{1}.$
Explanation:
$D$ represents the total number of ways of selecting two pieces of yellow fruit.
$N$ represents the total number of ways of selecting one yellow apple and one yellow banana.
Addendum
Responding to comment question:
The more formal approach is to use Bayes Theorem.
That is, given events $R,S$ you have that
$p(R|S) = \frac{p(RS)}{p(S)}.$
Let $(R)$ denote the event of 1 yellow apple and 1 yellow banana being chosen.
Let $(S)$ denote the event that two yellow pieces of fruit were chosen.
You want $p(R|S)$.
$p(RS) = [(5/18)\times (6/17)] + [(6/18)\times (5/17)].$
The above expression has two terms, which represents that you can choose either the yellow apple first, or the yellow banana first.
$p(RS)$ simplifies to $\frac{2 \times 5 \times 6}{18 \times 17}.$
$p(S) = 11/18 \times 10/17.$
So $\frac{p(RS)}{p(S)} = \frac{2 \times 5 \times 6}{11 \times 10}.$
Best Answer
Hint: Let $f(n)$ be the number of ways to eat $n$ pieces of fruit ending with an apple. Because we get stuck on apples, let $g(n)$ be the number of ways to eat $n$ pieces of fruit ending with a banana. There are also $g(n)$ ways to eat $n$ pieces of fruit and ending with an orange. Let $h(n)$ be the number of ways to eat $n$ pieces of fruit ending with strawberries. You should be able to write coupled recurrences for $g(n),h(n)$. Then the total number of ways of eating $n$ pieces of fruit is $f(n)+2g(n)+h(n)$
An apple can go after anything, so $f(n)=f(n-1)+2g(n-1)+h(n)$. A banana can go after an orange or a strawberry, so $g(n)=g(n-1)+h(n-1)$. A strawberry can go after anything but an apple, so $h(n)=2g(n-1)+h(n-1)$
A spreadsheet with the result is below
