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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!

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  • $\begingroup$ Does this work? $\frac{1.10.10+1.9.9+1.8.8}{1000}=\frac{49}{200}$? $\endgroup$ Commented Jun 2, 2022 at 19:50
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    $\begingroup$ the answer is 17/24 $\endgroup$ Commented Jun 2, 2022 at 20:28
  • $\begingroup$ To the OP: See the Addendum-2 at the end of my answer. $\endgroup$ Commented Jun 2, 2022 at 23:35
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    $\begingroup$ there are 10 cards and a number from 1 to 10 is written on each card --> Question from a non-native speaker: Shouldn't that be "the numbers from 1 to 10 are written on the cards". Because IMHO as the question is stated now it could also be that you have e.g. 10 cards with only the number 8 on them. And will this lead to a completely different result? $\endgroup$ Commented Jun 3, 2022 at 7:28
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    $\begingroup$ @OcasoProtal Technically yes, in reality no. English speaking is complicated and often bizarre. You have touched on the distinction between a denotation (i.e. the technical meaning of the words used in the phrase) and a connotation (i.e. the meaning inferred by others, upon reading the words in the phrase). In the setting of this problem, it was generally assumed that each card had a distinct element from the set $\{1,2,\cdots,10\}.$ Therefore, the (imprecise) communication was in fact effective. $\endgroup$ Commented Jun 3, 2022 at 9:09

3 Answers 3

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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?

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  • $\begingroup$ So my approach won't work because I am saying that no matter what the first card is a card that I need, when in reality it's not that simple? How could I have fixed my way of solving? Btw, I didn't even think about the complementary stuff. THANK YOU! I guess if you want to find P(A), you can always just 1-P(B) to get P(A) (If P(B) is the compliment) Will remember it for sure! $\endgroup$ Commented Jun 2, 2022 at 20:31
  • $\begingroup$ Exactly, using complements is frequently very useful! In terms of your method, you are actually very close. Notice that if you multiply your answer by 3, you get the correct result. The reason for this is that you correctly identified the relevant probabilities, but didn't take into account that for example, $1,A,A$ could also occur as $A,1,A$ and $A,A,1$. The order matters (which is what I was trying to get at in my answer) $\endgroup$
    – masiewpao
    Commented Jun 2, 2022 at 22:33
  • $\begingroup$ Is that 3 supposed to come from permutations? Where does that 3 come from? $\endgroup$ Commented Jun 2, 2022 at 22:37
  • $\begingroup$ @TizzleRizzle yes. For example, you identified the probability of the situation with the first card being a $1$. But what if instead the second card was a $1$? Or the third? multiplying by three, you cover all (mutually exclusive) scenarios $\endgroup$
    – masiewpao
    Commented Jun 2, 2022 at 22:39
  • $\begingroup$ so by multiplying by 3, what is happening to each of the cards individually? is the 3 coming from 3 cards total or something? $\endgroup$ Commented Jun 2, 2022 at 22:47
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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)$

$=1-(7/10) \cdot (6/9) \cdot (5/8)$

$=17/24$

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    $\begingroup$ Thank you! Really good explanation that I understood right away! $\endgroup$ Commented Jun 2, 2022 at 20:32
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Addendum-2 added to respond to the comment of masiewpao


An alternative is to express the probability combinatorically as

$$1 - \frac{\binom{7}{3}}{\binom{10}{3}} = 1 - \frac{35}{120} = \frac{17}{24}.\tag1 $$

In (1) above, when computing the RHS fraction, you have to be consistent between the numerator and denominator re whether order of selection is deemed important. For convenience, I used Combinations, which is equivalent to saying that in both the numerator and denominator, order of selection was deemed unimportant.

So, the RHS numerator represents all of the ways of choosing $3$ items, sampling without replacement, from the set $\{4,5,6,7,8,9,10\}$, where order of selection is deemed unimportant.

Since the fraction represents the probability that all $3$ numbers are above $3$, you take the complementary probability (i.e $1$ minus the fraction) to determine the probability that at least one of the cards was below a $4$.


Addendum
For what it's worth, the approach taken by the OP (i.e. original poster), although not recommended, is workable.

In order to implement his direct approach of summing probabilities, you have to identify all possible satisfactory mutually exclusive events, and add them up.

A satisfactory event is if there is either $1$ card below a $4$, $2$ cards below a $4$, or $3$ cards below a $4$.


$\underline{\text{Case 1: 1 Card below a 4}}$

First, I will assume that the first card drawn was the lowest card. Then, I will apply the scalar of $(3)$ to adjust for the fact that any one of the $3$ cards might have been the low card drawn.

The probability that the 1st card is $3$ or less is $\displaystyle \frac{3}{10}.$

Then, the probability that the 2nd card is $4$ or greater is $~\displaystyle \frac{7}{9}. ~$ This is because after the first card is drawn, there are $9$ cards left, $7$ of which are $4$ or greater.

Similarly, the probability that the 3rd card is also $4$ or greater will be $~\displaystyle \frac{6}{8}$.

Putting this all together, the probability of Case 1 occurring is

$$3 \times \frac{3}{10} \times \frac{7}{9} \times \frac{6}{8} = \frac{378}{720}. \tag2 $$


$\underline{\text{Case 2: 2 Cards below a 4}}$

First, I will assume that the first card drawn was the highest card. Then, I will apply the scalar of $(3)$ to adjust for the fact that any one of the $3$ cards might have been the high card drawn.

The probability that the 1st card is $4$ or more is $\displaystyle \frac{7}{10}.$

Then, the probability that the 2nd card is $3$ or less is $~\displaystyle \frac{3}{9}. ~$ This is because after the first card is drawn, there are $9$ cards left, $3$ of which are $3$ or less.

Similarly, the probability that the 3rd card is also $3$ or less will be $~\displaystyle \frac{2}{8}$.

Putting this all together, the probability of Case 2 occurring is

$$3 \times \frac{7}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{126}{720}. \tag3 $$


$\underline{\text{Case 3: 3 Cards below a 4}}$

The probability that the 1st card is $3$ or less is $\displaystyle \frac{3}{10}.$

Then, the probability that the 2nd card is $3$ or less is $~\displaystyle \frac{2}{9}. ~$ This is because after the first card is drawn, there are $9$ cards left, $2$ of which are $3$ or less.

Similarly, the probability that the 3rd card is also $3$ or less will be $~\displaystyle \frac{1}{8}$.

Putting this all together, the probability of Case 3 occurring is

$$\frac{3}{10} \times \frac{2}{9} \times \frac{1}{8} = \frac{6}{720}. \tag3 $$


Final computation:

$$\frac{378}{720} + \frac{126}{720} + \frac{6}{720} = \frac{510}{720} = \frac{17}{24}.$$


Addendum-2
Addendum-2 added to respond to the comment of masiewpao.

His comment indicates that my Addendum is overly complicated and that the alternative (simpler) approach that the OP (i.e. original poster) was going for is doable.

I agree. When I looked at the original posting, I didn't spend that much time trying to dissect the OP's intent. So, the following represents how the OP's approach would be implemented.

First, examine what the OP is doing. He is considering the following mutually exclusive cases:

  • The first card is a $1$.
    Probability is $\displaystyle\frac{1}{10}.$

  • The first card is a $2$, and the other two cards are both above a $1$.
    Probability is $\displaystyle\frac{1}{10} \times \frac{8}{9} \times \frac{7}{8} = \frac{56}{720}.$

  • The first card is a $3$, and the other two cards are both above a $2$.
    Probability is $\displaystyle\frac{1}{10} \times \frac{7}{9} \times \frac{6}{8} = \frac{42}{720}.$

Then, he reasoned that since these $3$ cases are mutually exclusive, they can be summed. The corresponding result is

$$\frac{1}{10} + \frac{56}{720} + \frac{42}{720} = \frac{170}{720}.$$

In fact, his analyis is exactly right, except for one subtle nuance. He assumed that the only way that he could get at least one of the cards to be $3$ or less is if the low card was the first card drawn.

In fact, the low card could be any one of the $3$ cards. Therefore, his computation of $~\displaystyle \frac{170}{720}~$ needs to be multiplied by $3$, which produces

$$\frac{170}{720} \times 3 = \frac{510}{720} = \frac{17}{24}.$$

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    $\begingroup$ This seems more complicated than what the OP was trying to do, he simply has to multiply his answer by three. And in saying that I mean it isn't a coincidence that the answer is a third of the right one; it falls out of the fact the OP didn't realise they had to account for the two extra permutations. $\endgroup$
    – masiewpao
    Commented Jun 2, 2022 at 22:34
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    $\begingroup$ @masiewpao : +1, nice catch, thanks. See my Addendum-2. $\endgroup$ Commented Jun 2, 2022 at 23:34

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