I have just started learning probability theory and came across this problem,
In each packet of Corn Flakes may be found a plastic bust of one of the last five Vice-Chancellors of Cambridge University, the probability that any given packet contains any specific Vice-Chancellor being $\frac{1}{5}$, independently of all other packets. Show that the probability that each of the last three Vice-Chancellors is obtained in a bulk purchase of six packets is $1 - 3 \big(\frac{4}{5}\big)^6 + 3 \big(\frac{3}{5}\big)^6 - \big(\frac{2}{5}\big)^6$
Now I understand how this is solved by considering each package as a seperate experiment and since their results (i.e which $V_i$ is inside) are independant, we have $p(V_1) = \big(\frac{1}{5}\big)^6$. Then we use the inclusion–exclusion principle.
However, I thought about it at first in a different way and I can't figure out why I am wrong. Our sample space $\Omega := \{\omega = (\omega^1,\ldots,\omega^6) : \omega^i \in \{1,\ldots, 5 \}\}$ where each $\omega^{i}$ denotes the result of opening $i$-th package. Because getting each $V_i$ in each package is equally probable, the probability of getting any $\omega \in \Omega$ should also be $\frac{1}{|\Omega|}$. Now since for our problem we don't care about the order (which $V_i$ is in which package) and we are also allowed for instance to get an $\omega$ like $(1,1,1,1,1,1)$, the cardinality should be ${N+n-1\choose n}$ where $N = 5$ and $n = 6$. Let $A_i \subset \Omega$ contain all the sequence where figure $V_i$ appears at least once. Then $$\mathbb{P}(A_3 \cap A_4 \cap A_5) = \sum_{|A_3 \cap A_4 \cap A_5|} \frac{1}{210} = |A_3 \cap A_4 \cap A_5| \cdot \frac{1}{210}$$ I was stuck calculating $|A_3 \cap A_4 \cap A_5|$ but each time I tried, the answer was still wrong. So I don't think my approach is correct to begin with.
Any help would be great! Thanks.
You may not care about the order, but the probabilities certainly do. Imagine that you toss a coin twice. Is the probability that you throw heads twice $\frac14$ or $\frac13$?
This is a question of how to apply (and how not to apply) the principle of indifference. The elementary equiprobable events here are the individual vice-chancellors being drawn in individual draws; so these are uniformly distributed. From this you can deduce by the multiplication principle that their ordered tuples are also uniformly distributed. There’s no reason to believe that the unordered tuples are uniformly distributed.