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Since we have 21 blocks in total. If we remove 4 blocks. We will end up with at most 5 complete sets left (since 21-4=17 and 17 can form at most 5 complete sets). We will also end up with at least 3 complete sets (remove a block from a distinct set each time).

Pr(5 sets remain) = 1 * 18/20 * 4/19 * 3/18

Pr(4 sets remain) = 1 * 18/20 * 15/19 * 6/18

Pr(3 sets remain) = 1 * 18/20 * 15/19 * 12/18

The 3 possible outcomes does not sum to probability 1. Please help

prob_try
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – CrSb0001 Oct 31 '23 at 15:31
  • There are two ways to have exactly five remaining sets. Either we wipe out one set completely and then take one from another set, or we take two from each of two sets. So...that probability ought to be a sum. And so on. – lulu Oct 31 '23 at 15:45

4 Answers4

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We need to count the reduced sample space for $3/4/5$ complete sets (i.e. remove from $4/3/2$ sets ), and compute the probabilities.
I presume each block in a set as distinct

  • $3$ intact, remove one each from $4$ sets.
    $\binom74(\binom31)^4 =2835$

  • $4$ intact, remove from $3$ sets $(2-1-1)$: Choose $3$ sets, choose set which will have $2$ removed, choose blocks, giving $\binom73\binom31\binom32(\binom31)^2 = 2835$

  • $5$ intact, remove from $2$ sets, $(3-1/1-3) \;or\;(2-2)$ in similar fashion, $\binom72[(2*\binom33\binom31) +(\binom32)^2] = 315$

Restricted sample space $= 2835+2835+315= 5985$

P($3$ sets intact|$4$ blocks removed $=\frac{2835}{5985} =\frac9{19}$

P($4$ sets intact|$4$ blocks removed $=\frac{2835}{5985} =\frac9{19}$

P($5$ sets intact|$4$ blocks removed = $\frac{315}{5985} =\frac1{19}$

true blue anil
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$\mathtt{ "Slot\; Method"\; Answer}$

I see that you attempted to use what I call the "slot method" which, as you can see here can be a huge shortcut. However, unless you are careful, in more complex situations, you can miss nuances.

I have already given a conventional answer, so here I shall be brief, following your answer with the necessary refinements.

P($3$ sets remain) $= 1 * \frac{18}{20}\frac{15}{19}\frac{12}{18} = \frac9{19}$ $\mathtt{(Simple\; Version\; of\; Slot\; Method)}$

P($4$ sets remain) $= \binom42 *1*\frac2{20}\frac{18}{19}\frac{15}{18} = \frac9{19}$ $\mathtt{(We\; first\; need\; to\; choose\; which\; two\; will\; be\; together)}$

$\mathtt{Five\; sets\; remaining\; can\; discard\; 2-2\; or\; 3-1}$

P($5$ sets remain) $= 3*1*\frac {2}{20}\frac{18}{19}\frac{2}{18} +4*1*\frac{2}{20}\frac{1}{19}\frac{18}{18} = \frac{1}{19}$

ADDED: Explanation of multiplication factors

You can consider the blocks as forming unlabeled teams (except where they become labeled by different sizes), thus

  • $2-1-1:\; \binom42\binom21\binom11\div 2! =6$
  • $2-2:\; \binom42\binom22\div2! =6$
  • $3-1:\; \binom43\binom11 = 4\;$ (No divisor needed as labeled by size)
  • By the same process, you should understand why the $1-1-1-1$ configuration doesn't need a multiplication factor (= has a m.f. of $1$), as $\binom41\binom31\binom21\binom11\div4! = 1$

I have explained the multiplication factors at length, but they are actually quite simple, (break into teams the normal way and divide by $r!$ where $r$ unlabeled teams are there)

true blue anil
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  • Hi thanks for the explanation. Would you mind elaborating a little more on the factors portion? What does AB, AC, AD represent here. – prob_try Nov 01 '23 at 15:29
  • We are splitting 2-2, so AB-CD, AC-BD, AD-BC, only three splits are possible. – true blue anil Nov 01 '23 at 15:44
  • I have added an explanation of all the multiplication factors, which you need to bear in mind while using the "slot method" It should be crystal clear now. – true blue anil Nov 02 '23 at 20:31
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If $n$ complete sets are remaining, there are ${{7}\choose{n}}$ ways to select which sets those are. There are then $17-3n$ blocks remaining, to distribute among the $7-n$ incomplete sets.

  • $n=5$. ${{7}\choose{5}}=21$ ways to choose the complete sets. $2$ blocks remaining to assign to the $2$ incomplete sets... ${{6}\choose{2}}=15$ ways to do this, or $21\times 15=315$ total options.

  • $n=4$. ${{7}\choose{4}}=35$ ways to choose the complete sets. $5$ blocks remaining to assign to the $3$ incomplete sets; must be $2-2-1$. So $3^4=81$ ways to do this, or $35\times 81 = 2835$ total options.

  • $n=3$. ${{7}\choose{3}}=35$ ways to choose the complete sets. $8$ blocks remaining to assign to the $4$ incomplete sets; must be $2-2-2-2$. So $3^4=81$ ways to do this, or (again) $35\times 81 = 2835$ total options.

The total number of options should be ${{21}\choose{4}}=5985$, and indeed $315+2835+2835=5985$. So the final probabilities for $n=3,4,5$ are $9/19$, $9/19$, and $1/19$, respectively.

mjqxxxx
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  • Hi, thanks for your explanation. For n=5, understand that 6 C 2 = 6 C 4, but logic wise shouldn't we have 4 blocks remaining to assign instead of 2? We are picking the 5 "complete" sets out of 7 total sets, and for the remaining 2 "incomplete" sets, we pick 4 blocks out of a total of 2*3=6 blocks from the "incomplete" sets. – prob_try Nov 01 '23 at 15:42
  • The "blocks remaining to assign" are the blocks remaining (i.e., not removed) in the incomplete sets. For $n=5$, after removal there are $15$ blocks in the complete sets and $2$ in the incomplete sets. Of course, you can approach the calculation the other way (decide which blocks in the incomplete sets were removed), and you'll get the same result. – mjqxxxx Nov 01 '23 at 15:56
  • Apologies just want to clarify again: for n=4, we have 3 non-complete sets and we are arranging the 5 blocks remaining, is the logic like: (3 blocks choose 1) ^3 * (3 sets choose 1 to place the last remaining block) = 3^4?

    And for n=3, we have 4 non-complete sets, so we are picking (3 blocks choose 2) ^ 4 = 3^4. 4 comes from the fact that we have 4 types of blocks to allocate?

     – prob_try Nov 01 '23 at 16:08
  • We are choosing which blocks didn't get removed. For $n=4$, there are $5$ blocks remaining among the $3$ incomplete sets. We must have less than three in each of those sets (or they wouldn't be incomplete); the only option is $2-2-1$, where the single block can be in any of those incomplete sets. So ${{3}\choose{2}}^2 \cdot {{3}\choose{1}}=3^3$ ways to choose which blocks remain, times $3$ ways to choose which set gets the single block, is $3^4$. For $n=3$, there are $8$ blocks remaining among the $4$ incomplete sets. We must have exactly two in each set. Again ${{3}\choose{2}}^4=3^4$. – mjqxxxx Nov 01 '23 at 16:56
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An alternate approach using generalised inclusion/exclusion:

Let's say the final arrangement has "Property $i$" if no blocks have been removed from set $i$, for $1 \le i \le 7$, and define $S_j$ to be the total probability of the arrangements with $j$ of the properties, for $0 \le j \le 5$. Then $$S_j = \binom{7}{j} \binom{21-3j}{4} \bigg/ \binom{21}{4}$$ We want to find the probability of an arrangement's having exactly $k$ of the properties, i.e. exactly $k$ intact blocks, for $3 \le k \le 5$. There are (at least) two ways to proceed: we can use generalised inclusion/exclusion, as described in Generalised inclusion/exclusion principle, or we can use a generating function. I will illustrate the generating function method here and leave the equivalent generalised inclusion/exclusion method for the reader.

Let $f(x)$ be the generating function for $S_j$, i.e. $$\begin{align} f(x) &= \sum_{j=0}^5 S_j x^j \\ &= 1+\frac{68 x}{19}+\frac{91 x^2}{19}+\frac{55 x^3}{19}+\frac{14 x^4}{19}+\frac{x^5}{19} \end{align}$$ Now if we let $g(x) = f(x-1)$, then $g(x)$ is the generating function for the exact probabilities; see, for example, Section 4.2, "A generatingfunctionological view of the sieve method" in generatingfunctionology, Second Edition, by Herbert S. Wilf (which is available online).

After expanding $f(x-1)$, we find $$g(x) = \frac{9 x^3}{19}+\frac{9 x^4}{19}+\frac{x^5}{19}$$ where we see the desired probabilities.

awkward
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