Probablity of a card being less than or equal to 3 In a box, there are 10 cards and a number from 1 to 10 is written on each card. When three cards from the box are randomly taken at a time, we define X,Y, and Z according to three numbers in ascending order. The probablity that X is less than or equal to 3 is:
I tried writing out what the probablity of three situations would be where A is anything.
$$1AA = 1/10 * 1 * 1$$
$$2AA (excluding 1) = 1/10 * 8/9 * 7/8$$
$$3AA (excluding 2 and 1)=  1/10 * 7/9 * 6/8$$
After adding all of these up I came no where near the answer: $17/24$or($85/120$also works)
Where am I going wrong with this? Also, how do I solve it?
I thought about permutations, and how many different ways we could draw these cards, but it seems like the cards have to be in a strict order (ascending) so even if we draw the cards out of order, they will be put in order, so everything is just multiplied by 1, since there are no permuations (or so I think)
I also thought about what if this is just asking, of a random set of three cards, what is the chance that x is less than 3? and thought
XYZ, X has a 3/10 chance to be 3 or less. However, after that I got lost on how I should multiply 3/10, since the next two numbers in that sequence are fully dependent on the first number.
All help is appreciated! thank you!
 A: Here is a way to think of the problem statement: The question asks that at least one of the three cards drawn is no bigger than a 3. Alternatively, we can consider the case where all three cards are in fact bigger than a 3. Note that if we can calculate the probability of this event we are done. This is because this event is the complement of the one we are interested in (so the final probability is one minus the probability of all three cards being greater than 3).
But this is isn't too hard to see: The probability of the first card being strictly larger than a 3 is $\frac{7}{10}$. This is because of the ten cards, there are seven cards greater than a 3: $4,5,6,7,8,9,10$. For the second card, the probability it is greater than a 3 is $\frac{6}{9}$. This is because we assume the first card is one of $4,5,6,7,8,9,10$, and that this is removed from the pool of remaining cards. The exact same logic gives us the probability that the third cared is greater than a 3 is $\frac{5}{8}$.
So our answer is $1-\big(\frac{7}{10}\cdot\frac{6}{9}\cdot\frac{5}{8}\big) = \frac{17}{24}$ .
At a first glance an issue with your approach: You are assuming that the card with the smallest value occurs in the first card you draw. I think I see why you thought this, because the question is phrased in a slightly confusing way. The question is not saying X,Y,Z correspond to the first, second and third cards respectively. Instead, it is saying that of the three cards you draw, assign the card with the smallest value to X, the card with the 'mid' value to Y, and the card with the largest value to Z. Do you see now why your approach won't work?
A: As the problem states, we have 10 cards labeled 1 through 10. Formally we can describe your problem as finding finding $\mathbb{P}(\min(X, Y, Z) \leq 3)$
where X, Y and Z are the numbered cards pulled without replacement. Then we can perform the following manipulation using the complement rule:
$\mathbb{P}(\min(X, Y, Z) \leq 3) = 1-\mathbb{P}(\min(X, Y, Z) > 3)$.
We can then simplify this by observing that if the $\min(X,Y,Z) > 3$, then X,Y,Z must all be greater than 3. So
$1-\mathbb{P}(\min(X, Y, Z) > 3)$
= $1-\mathbb{P}(X>3, Y>3, Z > 3)$
= $1-\mathbb{P}(X>3)$$\cdot \mathbb{P}(Y>3|X > 3) \cdot \mathbb{P}(Z>3|X > 3,Y>3)$
$=1-(7/10) \cdot (6/9) \cdot (5/8)$
$=17/24$
