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This is from a book.

enter image description here

My answer was 35/210 using non-negative integral solutions method approach.

The only thing i can reason is that the books method counts every packet as seperate.

Which answer is correct.

Edit : My Solution :

X1 + X2 + X3 + X4 + X5 = 6

The answer is the number of solution to this such that (X1 >=0 ,X2>=0 ,X3>=1 ,X4 >=1 , X5 >=1), which was 35

And the total probability is : The number of solutions to this such that (X1 >=0 ,X2>=0 ,X3>=0 ,X4 >=0 , X5 >=0). which was 210.

pppp_prs
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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. – Community Feb 09 '23 at 06:45
  • Please edit in the calculations of your approach – true blue anil Feb 09 '23 at 07:00
  • How can you fail to get all three? How many ways can you not get two and how many ways can you not get one? What are those probabilities? Try to answer those questions while looking at the answer. – John Douma Feb 09 '23 at 07:23
  • i can definitely think along these lines. I want to know what is the problem with my approach – pppp_prs Feb 09 '23 at 07:31
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    Consider the case where you get three of the 1st VC and three of the 2nd. That just counts as one of your $210$ solutions, $x_1=x_2=3$, $x_3=x_4=x_5=0$. But in fact it should count as $20$ outcomes, since there are $20$ ways to choose which three of the packets has the 1st VC. So, your $210$ is a massive undercount of the total number of outcomes. – Gerry Myerson Feb 09 '23 at 08:20
  • @GerryMyerson You are assuming that the 6 packets are different and not identical. However to me it is not clear from above framing of question if we can assume this or is it? – pppp_prs Feb 09 '23 at 12:19
  • Even if I throw two identical dice, the probability of throwing two ones is not the same as the probability of throwing a one and a two. The dice are identical so 1+2 is considered the same outcome as 2+1, but we still have to count both options when we work out the probabilities. – Jaap Scherphuis Feb 09 '23 at 12:38
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    Suppose we flip 10 identical coins. If we used your method, we'd have one variable for the number of heads, another for the number of tails, adding to 10. There would be 11 solutions, and you would say that the probability of having only heads was 1/11, which is obviously wrong. Those 11 solutions do not represent equally probable outcomes, even though the coins are identical. – Jaap Scherphuis Feb 09 '23 at 12:52
  • The number of non-negative solutions of $X_1 + X_2 + X_3 + X_4 + X_5 = 6$ is a combinatorics problem. As a general rule, never assume that combinatorics will give you a set of equiprobable events in a probability problem. Some formulas from combinatorics (such as the binomial coefficients) do show up in probability as well, but you have to derive them using methods of probability. We should not need to show where your approach is broken; you should be able to show why it must work, if it works. – David K Feb 09 '23 at 17:14
  • The words "independently of all other packets" tell us that if we take two packets from the bulk purchase, mark one "A" and the other "B", there is exactly a $1/25$ probability that packets A and B both contain busts of Stephen Toope. Only a model that produces this probability of that event is valid for this problem. How does your approach give us this probability? – David K Feb 09 '23 at 17:20
  • Please do not rely on pictures of text – Shaun Apr 28 '23 at 23:29
  • Which book are you referring to? – Shaun Apr 28 '23 at 23:29

1 Answers1

2

To come to your doubt

  • Using stars and bars is totally inappropriate here, because the alternatives thrown up by it are not equiprobable Just commonsense would tell you that getting all throws in a single "box", so to say, is much more unlikely than a more even distribution.

  • Since a bulk purchase of $6$ boxes has been made, the situation can be correctly modeled as throwing together a bunch of six five faced dice (which creates an equiprobable sample space), and the solution is based on this realisation.

The solution uses inclusion-exclusion, counting

[All possible results]
$-$ [At least $1$ from the last three missing]
$+$ [At least $2$ from the last three missing]
$-$ [All $3$ from the last three missing]

$$ Pr = 1 - \frac{\binom31 4^6 +\binom32 3^6 - \binom33 2^6}{5^6}$$

PS

It is clear that each of the packets have an equal probability of $\Large\frac15$ of having the bust of any of the $5$ vice-chancellors.

To understand in greater detail why a die throw model is appropriate here and a "stars and bars" model is not, you can see a detailed worked out explanatory example in my answer here

true blue anil
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  • Please explain "because the alternatives thrown up by it are not equiprobable Just commonsense would tell you that getting all throws in a single "box", so to say, is much more unlikely than a more even distribution." Its not clear to me intuitively. – pppp_prs Feb 09 '23 at 12:29
  • @pppp_prs: I have added a PS with link in the main body itself. – true blue anil Feb 09 '23 at 16:56
  • @pppp_prs: Are you still having doubts ? If so, what ? – true blue anil Feb 16 '23 at 17:10