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A box of cereal may contain one of 5 figurines of the last 5 Roman Emperors, each with probability $\frac{1}{5}$

What is the probability of having the last 3 Emperors in a bulk purchase of 20 boxes ?

I understand that I should use the Inclusion-exclusion principle but I am not sure how ...

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    if A,B,C are events that an emperor 1,2,3 is missing, then prob any are missing is P(A) + P(B) + P(C) + ...... etc If you know the prob that any of the 3 are missing - then ..what? – 2017-02-21
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    So I think maybe P(receiving #1 and receiving #2 and receiving #3) =1-P(#1 missing or #2 missing or #3 missing) ? – 2017-02-21
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    you need to work out the probability of NOT receiving any one of the 3 (the compliment of this is the probability you require) - but you can't just do $3 \times (0.8)^{20}$ of course – 2017-02-21
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    yes you are right there – 2017-02-21
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    that is one thing I don't understand - why does the inclusion-exclusion give a different answer from the simple multiplication in your comment ?- after all receiving/not receiving a particular emperor are all independent events ? – 2017-02-21
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    Presumably each box contains exactly one figurine – 2017-02-21
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    @henry : yes exactly 1 figurine per nox – 2017-02-21
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    P(A) + P(B) + P(C) would 'double count' case where A and B occurred - the simplest example is P(A or B) = P(A) + P(B) - P(A and B) – 2017-02-21

1 Answers 1

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  • The probability all $20$ boxes are missing a particular figurine is $\left(\frac45\right)^{20}$
  • The probability all $20$ boxes are missing two particular figurines is $\left(\frac35\right)^{20}$
  • The probability all $20$ boxes are missing three particular figurines is $\left(\frac25\right)^{20}$

So the probability that three particular figurines are in the $20$ boxes is $$1-3\times \left(\frac45\right)^{20} +3 \times \left(\frac35\right)^{20} - \left(\frac25\right)^{20} \approx 0.96552$$

This is not the same as the simple multiplication $\left(1-\left(\frac45\right)^{20}\right)^3 \approx 0.96581$ though it is not far away because $\left(1-\left(\frac45\right)^{20}\right)^3 = 1-3\times \left(\frac45\right)^{20} +3 \times \left(\frac45\right)^{40} - \left(\frac45\right)^{60}$ and the two right hand terms in each expression are small