# Probability that at least two equal numbers are extracted

Consider an urn containing 90 balls numbered from 1 to 90, plus 3 balls attached with 3 distinct (known) numbers still from 1 to 90, say 1,2,3. I'm trying to find the probability that at least two equal numbers are extracted after 5 extractions without replacement. I've tried two approaches which give different answers, and I don't really know which is correct (mistakes aside).

Firstly, with $\Omega_0=\{1,2,\cdots,90,1a,2a,3a \}$ I would define as a sample space $\Omega=\{\omega=(\omega_1,\omega_2,\omega_3,\omega_4,\omega_5)\in\Omega_0^5\},$ as a $\sigma$-algebra $F=P(\Omega)$ and as a probability measure $\mathbb{P}$ such that $\mathbb{P}(\{\omega\})= 1/ \binom {93}{5} .$

Let $A$ be the event of two equal numbers being extracted. My first approach is to consider $A^c=\{\omega \in \Omega :\omega_i \ne\omega_j \text{ for } i \ne j\};$ I think $\lvert A^c\rvert =\binom{90}{5}$, which gives $\mathbb{P}(A)=1-\frac{88\cdot87\cdot86}{93\cdot92\cdot91}.$

Second approach: $\mathbb{P}(A)=\mathbb{P}(\text{two 1's" or }\text{two 2's" or }\text{two 3's}")=3\mathbb{P}(\text{two 1's}") -3\mathbb{P}(\text{two 1's or two 2's}")$. Then from $$\mathbb{P}(\text{two 1's}")=\frac{\binom{2}{2}\binom{91}{3}}{{93}\choose{5}}$$ and $$\mathbb{P}(\text{two 1's and two 2's}")=\frac{\binom{2}{2} {2\choose2} {{89}\choose1}}{{93}\choose{5}}$$ I proceed to the result but it doesn't equal what I got earlier. Where do I go wrong?

• I'm not succeeding in perfect formatting, I would be really glad if someone could fix the body Commented Apr 7, 2017 at 21:46
• How many balls are drawn? It looks like five are drawn without replacement, so the only duplicates you can get are one of the $a$s and the matching ball without the $a$. Is that right? I don't understand the logic of the first approach. Your second approach makes good sense to me. Commented Apr 7, 2017 at 22:59
• @Max: Thank you for your edit! Commented Apr 7, 2017 at 23:18
• @RossMillikan I'm sorry, sir, I forgot to mention those details but you got everything right. About the second approach, is it correct that $\lvert A^c\rvert =\binom{90}{5}$? Commented Apr 7, 2017 at 23:25
• @Richard Sure thing. I realized that it wanted the braces {n \choose k} . Compare $100\choose 1{9 \choose 2}$ 100\choose 1{9 \choose 2}, ${100\choose 1}{9 \choose 2}$ {100\choose 1}{9 \choose 2}, and $100\choose 1 9 \choose 2$ 100\choose 1 9 \choose 2. You can also use \binom{n}{k}` $\binom{n}{k}$.
– Em.
Commented Apr 7, 2017 at 23:40

First, if you define the sample space $\Omega=\{\omega=(\omega_1,\omega_2,\omega_3,\omega_4,\omega_5)\in\Omega_0^5\}$, you take order of element into account. This contradicts to $|\Omega|=\binom{93}{5}$.

The valid elementary event for $\Omega$ without ordering should be $$\{\omega_1,\omega_2,\omega_3,\omega_4,\omega_5\}, \omega_i\in\Omega_0.$$

Next, $|A^c|$ does not equal to $\binom{90}{5}$. Why? In $\binom{90}{5}$ you counted only fives with all elements taken from $\{1,2,3,\ldots,90\}$. But the fives $\{1a,2,3,6,7\}$, $\{1a,2a,3,5,6\}$, $\{1a,2a,3a,6,7\}$ also belong to $A^c$. So you need to add the number of those fives.

The number of firsk kind fives is the nouber of ways we choose one element from $\{1a,2a,3a\}$ multiplied by the number of ways to choose $4$ numbers from the set $\{1,2,\ldots,90\}$ without element which is paired to already choosen: $\binom{3}{1}\binom{90-1}{4}$. And $$|A^c|=\binom{90}{5}+\binom{3}{1}\binom{89}{4}+\binom{3}{2}\binom{88}{3}+\binom{3}{3}\binom{87}{2}.$$

A nice problem. The overall number of combinations is easy to find $${93\choose 5}$$

Now the number of favorable combinations. There’s a whole lot to account for in here, and that’s what is interesting here.

Type I of possible favorable outcomes. One of the "attached" three balls is drawn with its counterpart with the same number out of $$90$$. The number of combinations: $$3$$ (because we have three choices of extra balls) times $${89\choose 3}$$ (because we have to exclude one favorable ball out of 90 to make sure our outcome is favorable. Then our favorable-outcome pair is combined with three balls out of $$89$$: $$= 3\cdot{89\choose 3}$$

Type II of possible favorable outcomes. Two of the "attached" three balls are drawn. It can be done in three ways $${3\choose 2}=3$$. So we will have two multiply each of the subtype outcomes (A, B) by three.

Type II Subtype A outcomes. Only one favorable ball (with the same number as is one of the two of the drawn "attached" balls) is drawn out of 90 to make our combination favorable. Now, out of the two of the "attached balls we drew, we pair either the first or the second with its respective counterpart from the "90 urn". So therefore we have $$2\cdot{88\choose 2}$$. That is we exclude one favorable ball out of $$90$$ to make sure our outcome is favorable by adding it to the two "attached" balls we drew. And we exclude one more favorable ball out of the rest of $$89$$ to prevent double count (see TypeII Subtype B right below) and get the number $$88$$. So we have this favorable triplet or $$3$$ balls to combine with the rest $${88\choose 2}.\quad$$ So $$2\cdot{88\choose 2}$$

Type II Subtype B outcomes. Two favorable balls (with the same numbers on them as our drawn "attached" balls) are drawn out of $$90$$ to make our combinations favorable. It's gonna be the count of two favorable pairs (two "attached" balls + its two respective counterparts from the "main" urn) 2+2, so to speak. So (we excluded two favorable balls out of $$90$$): $${88\choose 1}$$ Now we add subtypes I and II of Type II, remembering to multiply them by three (see the explanations in the beginning of Type II): $$3\cdot\bigg [2\cdot{88\choose 2}+{88\choose 1}\bigg]$$

Type III of possible favorable outcomes. All the three "attached" balls are drawn. To make the favorable combinations:

one of them (Type III subtype A) is paired with one of the $$90$$. It is three possible "quartets" of balls to be combined with $$87$$ balls. Again we excluded two more favorable counterparts ($$89-2=87$$) to prevent double count (it's accounted for in Type III subtype B): $$3\cdot{87\choose 1}$$

two of them (Type III subtype B) are paired with the two balls (with the same numbers) out of the $$90$$. We have only three possible "quintets": $$3\cdot{87\choose 0}=3$$ (again we used $$87$$ here to prevent double count, although it's just three possibilities for quintets anyway)

three (all) of them (Type III subtype C is not possible) can't be paired with the three out of $$90$$ 'cause we'll then have six balls. Not possible.

Now we add up all type III possibilities: $$3\cdot{87\choose 1}+3\cdot{87\choose 0}$$

Now, at long last, we add up all the favorable combinations and divide them by $${93\choose 5}$$:

$$\frac{3\cdot{89\choose 3}+3\cdot\bigg [2\cdot{88\choose 2}+{88\choose 1}\bigg]+3\cdot{87\choose 1}+3\cdot{87\choose 0}\,}{{93\choose 5}}≈0.007$$

At first the number looks smallish but in fact it should be small. Applying the theory of probability leads to the same smallish result. This was the simplest solution with the precise formula. The solution could have been much shorter, but I thought extra clarity would not go amiss. Such problems might be very, very confusing. So I elaborated on details.

Hope it was helpful and more or less clear.

• Thank you very much! However it appears your value for $\lvert A^c\rvert$ differs from that of the other answer, doesn't it? Commented Apr 8, 2017 at 12:23
• No, it does not. The numberin numerator is ecactly $|A|=364188$ as in your right solution and as $|\Omega|-|A^c|$ in my solution.
– NCh
Commented Apr 8, 2017 at 14:26
• @NCh I see, I must have made some typo in checking on Wolfram. Thank you too! Commented Apr 8, 2017 at 14:30
• Oh, how many typos I've made in previous comment... Sorry for that. I cannot edit it now.
– NCh
Commented Apr 8, 2017 at 14:40