Probability of drawing 3 certain cards?

Question

Say I have 10 cards uniformly distributed with probability 1/10. You now draw without replacement. Then I can use the binomial coefficient to compute the probability of drawing 3 cards where 1 certain card is not included: $$ \frac{\text{#ways to choose 3 cards from 9 cards}}{\text{#ways to choose 3 cards from 10 cards}} $$ This gives 0.7. Thus, it just gave the same result as my intuition told me. What it said was: The probability of drawing certain 3 cards is 30% so 1-30%=70%=0.7 which is the probability of not drawing this 3-card-combo which means you didn't draw that 1 card. However, why is my intuition matching the use of the binomial coefficient? Is the probability of drawing 3 cards really 30%? I mean I would say there is $\frac{1}{10*9*8}$ chance of drawing a certain 3-card-combo :/

Answer 1

It isn't that the probability of drawing $3$ cards is $30\%$. Suppose that the card to be excluded is the Ace of Spades. Then what your original question amounts to is:

Suppose that you divide up $10$ cards into $2$ groups, where one of the groups has $3$ cards and one of the groups has $7$ cards. What is the probability that the Ace of Spades is in the $7$ card group?

Answer:
The Ace of Spades has a $(1/10)$th probability of being any one of the $10$ cards. These disjoint probabilities are additive. So, the probability that the Ace of Spades is in the group of $3$ cards is $30\%$. Therefore, the probability that the Ace of Spades is not in the group of $3$ cards is $[100\% - 30\% = 70\%].$

The reason that this computation matches the result of the $~\displaystyle \frac{\binom{9}{3}}{\binom{10}{3}}~$ computation is that this alternative computation also happens to represent a valid approach to attacking the problem.