Given this table that shows the relationship between discounts on houses and condos and knowing that the supply distribution is 83% and 17% for houses and condos respectively I have to compute the probability that a home for sale is a house and that there have been given a 10% or greater discount.
So far, I have defined two events:
The probability that a home is a house P(B1) = 0.83 & the probability that a home is a condo P(B2)= 0.17.
I tried using the Multiplication Rule solve my problem:
0.12 * 0.83 = 9.96%
I am not sure this is correct, though, since I am having a hard time interpreting the information from the table.
Next question is: What is the probability that any given home was discounted more than 10%? I am completely lost here ...
You have done the right arithmetic so far, so I'll concentrate on the language and the notation for conditional probability, which seems to be causing some difficulty.
You have the table \begin{array}{cccc} \rlap{\text{discount on a house}} \\ \text{none} & 0\% \text{ to }10\% & > 10\% \\ 63\% & 25\% & 12\% \end{array}
The numbers in the bottom row add up to $100\%$, so it seems safe to assume that if you randomly pick a house that is for sale, there is a $12\%$ chance it is discounted more than $10\%$.
If you randomly pick any home, and then ask whether it is a house and the answer turns out to be "yes," then you can use the fact that you know this is a house to estimate the probability of a discount. That is, we again say there is a $12\%$ chance that the home is discounted more than $10\%$.
In the language of conditional probability,
To write this in mathematical notation, if $B_1$ is the event that the randomly selected home is a house and $D_{10}$ is the event that the randomly selected home is discounted more than $10\%$, then $P(D_{10} \mid B_1)$ is the probability a home for sale is discounted more than $10\%$ given that the home is a house, and we can write $$ P(D_{10} \mid B_1) = 0.12. $$
Now there is a rule for conditional probability that $$ P(A \text{ and } B) = P(A \mid B) \cdot P(B). $$
The application of this rule to your question is that $$ P(D_{10} \text{ and } B_1) = P(D_{10} \mid B) \cdot P(B) = 0.12 \cdot 0.83 = 0.0996, $$ which means that the probability a home for sale is discounted more than $10\%$ and is a house is $9.96\%$.
Now if you want the probability that a random home is discounted more than $10\%$, that is, you want $P(D_{10})$, since we already know every home for sale is either a house or a condo, the event $(B_1 \text{ or } B_2)$ is always true, and $$ D_{10} = D_{10} \text{ and } (B_1 \text{ or } B_2) = (D_{10} \text{ and } B_1) \text{ or } (D_{10} \text{ and } B_2). $$ That is, a home is discounted more that $10\%$ if and only if the home is discounted more that $10\%$ and is a house or the home is discounted more that $10\%$ and is a condo. Therefore $$ P(D_{10}) = P((D_{10} \text{ and } B_1) \text{ or } (D_{10} \text{ and } B_2)). $$
Is it clear that the event $(D_{10} \text{ and } B_1)$ and the event $(D_{10} \text{ and } B_2)$ are two mutually exclusive events (that is, they cannot both happen)? Then we can find the probability $P(D_{10})$ in terms of things we already know how to compute, such as $P(D_{10} \text{ and } B_1)$.