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What's the complement of the event $A$={not getting the number six when rolling a die 7 times}and why?

The complement is not (A). I'd say $A^c$= {getting the number six(only once) when rolling a die 7 times}

But what about $A^c$={getting at least one number six when rolling a die 7 times}?

Because the complement of getting at least one number six when rolling a die 7 times, is not getting the number six when rolling a die 7 times

EDIT: One die is rolled 7 times. What's the probability of getting the six possible numbers?

She started writting that $A_i$={not getting the $i$, $1 \leq i \leq 6$} in 7 rolls

$P(A_i) = {(\frac{5}{6})}^7$

$A^{\complement}_i$= {getting the $i$ in seven rolls}

$P(A^{\complement}_i)= 1-{(\frac{5}{6})}^7$

$P(\cap_{1}^{6} A^{\complement}_i)$ $= P(\cup A_i)^{\complement}$ (De Morgan's Law)

$= 1 - P(\cup_{1}^{6} A_i)$

Then she continues solving $P(\cup_{1}^{6} A_i)$ using the Inclusion-exclusion principle ...

$P(\cup_{1}^{6} A_i)$ =

$\sum_{i=1}^{6}$$P(A_i) - \sum_{i\neq j}P(A_i \cap A_j ) + \sum_{i\neq j \neq k}P(A_i \cap A_j \cap A_k)... (-1)^5 P(\cap_{1}^{6} A^{}_i) $

$= \binom{6}{1}(\frac{5}{6})^7 - \binom{6}{2}(\frac{4}{6})^7 + \binom{6}{3}(\frac{3}{6})^7 - \binom{6}{4}(\frac{2}{6})^7 + \binom{6}{5}(\frac{1}{6})^7 -\binom{6}{6} \cdot 0 = 0.945987$

$P(\cap_{1}^{6} A^{\complement}_i)$ $= 1- [\sum_{i=1}^{6} (-1)^{i+1} \binom{6}{i}\left ( [6-i]/6 \right )^7]= 0.054 $

The answer is right. I don't know what else could be wrong.

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  • $\begingroup$ The latter is the complement, not the former. $\endgroup$ – Don Thousand Apr 14 at 16:56
  • $\begingroup$ This is a doubt I had about a problem that the teacher solved. She said that the right answer was getting the number six only once. I can write the exercice if needed. $\endgroup$ – roy212 Apr 14 at 17:00
  • $\begingroup$ If you have presented the problem as intended, then your teacher is indeed wrong. $\endgroup$ – Don Thousand Apr 14 at 17:40
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$A^c=\{$not (not getting the number six when rolling a die 7 times) $\}=\{$ not (getting the number six exactly zero times when rolling a die 7 times)$\}=\{$getting another number of sixes than zero, when rolling a die 7 times$\}=\{$getting a six at least once when rolling a die 7 times$\}.$ Hence, your second answer is correct.

Edit: Your answer is right. Possibly an easier solution to the problem is $P=\frac{6\cdot\frac{7!}{2!}}{6^7}=\frac{35}{648}\approx5.4012\%.$ That is, because we need to devide the number of the "good" sequences by the number of all sequences. "Good" sequences of numbers (i. e. such sequences, that every number from 1 to 6 appears at least once) are permutations of one of the sequences $$(1,1,2,3,4,5,6);(2,1,2,3,4,5,6);(3,1,2,3,4,5,6);\\(4,1,2,3,4,5,6);(5,1,2,3,4,5,6);(6,1,2,3,4,5,6).$$ The sets of permutations are disjoint for each of those sequences. That is, because they all contain different seventh numbers. Each of those six sequences has $\frac{7!}{2!}$ permuations. Hence, the number of "good" sequences is $6\cdot\frac{7!}{2}.$ We have to divide that number by the number of all sequences that can be obtained by rolling a die 7 times, i. e. $6^7.$

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  • $\begingroup$ I posted the whole exercice $\endgroup$ – roy212 Apr 14 at 18:44

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