There are two questions and the approach seems to be different. a) Probability of choosing 3 queens from a deck of cards?
Ans:
4/52 * 3/51 * 2/50
However, if the question is how many diamonds would you expect to be drawn if 3 cards are randomly chosen, then the solution becomes
(13/52) * 3
In other words, it becomes
(13/52) + (13/52) + (13/52)
What is the key to understand when to add in probability questions and when to multiply?
On
If two events are statistically independent then we can multiply the probability of the two events occurring to get the probability of both events occurring. In the second part of your question you are using expectation algebra which is linear. That is to say $E(X+Y)=E(X)+E(Y)$ for all distributions or events $X$, $Y$. These two facts allow us to multiply in the first half and add in the second.
In the first part, the probability is calculated by counting the number of ways you can draw 3 queens (4 x 3 x 2) and dividing it by the total ways you can draw 3 cards (52 x 51 x 50)
the result is $\frac{4 \times 3 \times 2} {52 \times 51 \times 50} = \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50}$
Another way to think about it is by using conditional probability and understanding what the probabilities that are given mean:
The first is 4/52, this is the probability that the first card is a queen. let $A$ be the event that the 1st card is a queen.
so, $P(A) = \frac{4}{52}$ trivially because there are 4 queens out of 52.
Next, what does 3/51 mean? This is the probability that the 2nd card is also a queen as well. (after we know that the 1st is a queen). This is nothing but conditional probability. let $B$ be the event that the 2nd card is a queen.
By the formula for conditional probability, $$P(B|A) = \frac{P(B \cap A)}{P(A)}$$ So what is the probablity that the first 2 cards are queens?
This is $P(A\cap B) = P(A) \times P(B|A) = \frac{4}{52} \times \frac{3}{51}$
Great! What about the 3rd probability in your example? It is 2 / 50. What is this referring to? This is that the probablity that the 3rd card is also a queen (given that the first 2 are queens as well)
if $C$ is the event that the 3rd card is a queen, then $ \frac{2}{50} = P(C | A \cap B)$
Then the probability that all 3 cards are queens is $P(A\cap B\cap C) = P(A \cap B) \times P(C | A\cap B) = \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} $
Notice that we multiplied the probabilities when we where calculating the probability of the intersection of events like $P(A\cap B\cap C)$
We used conditional probabilities in the example because A, B, and C are not independent. (because B includes A) If the events where independent, we wouldn't need conditional probability because if A and B are independent then $P(B) = P(B|A)$,
and in this case then $P(A\cap\ B\cap C) = P(A) \times P(B) \times P(C)$
On the other hand, in your example about expectation, you have to realize that you are not even calculating a probability. To see that, assume the question was asking what is the expectation of drawing 5 diamonds instead of 3. In this case you will get 5 x (13 / 52) = 65 / 52 = 1.25 > 1
This is not even a probability value!
So if you want an example that shows the total probability being calculated by addition of probabilities, then consider this example:
What is the probability of drawing either a heart or a diamond. In this case you are calculating the probability of the union of the 2 events:
$A$: the card drawn is a heart where P(A) = 13 / 52
$B$: the card draw is a diamond where P(B) = 13 / 52
To calculate the desired probability, we are looking for $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Since A and B are independent, the $P(A\cap B)$ = 0
Therefore, $P(A \cup B) = P(A) + P(B) = 13 / 52 + 13 /52 = 26 / 52 = 0.5$
Notice we added probabilities when were were calculating the probabilities of union of event like $A\cup B$ and when A and B are independent.