Exercise 1.49

Question

A fair die is rolled $n$ times. What is the probability that at least 1 of the 6 values never appears?

fifthist x · Accepted Answer

Let $A_{i}$ be the event that $i$ is never rolled for $1 \leq i \leq 6$.
The event of interested then is $\bigcup\limits_{i=1}^{6}A_{i}.$

By inclusion-exclusion, 
$P(\bigcup\limits_{i=1}^{6}A_{i}) = \sum_{i=1}^{6}P(A_{i}) - 
\sum_{1 \leq i < j \leq 6}P(A_{i} \cap A_{j}) + 
\sum_{1 \leq i < j < k \leq 6}P(A_{i} \cap A_{j} \cap A_{k}) - 
\cdots - P(\bigcap\limits_{i=1}^{6}A_{i}).$

Now,

$P(A_{i}) = \frac{5^{n}}{6^{n}} = (\frac{5}{6})^{n}$

$P(A_{i} \cap A_{j}) = \frac{4^{n}}{6^{n}} = (\frac{4}{6})^{n}$

$P(A_{i} \cap A_{j} \cap A_{k}) = \frac{3^{n}}{6^{n}} = (\frac{3}{6})^{n}$

$P(A_{i} \cap A_{j} \cap A_{k} \cap A_{w}) = \frac{2^{n}}{6^{n}} = 
(\frac{2}{6})^{n}$

$P(A_{i} \cap A_{j} \cap A_{k} \cap A_{w} \cap A_{z}) = \frac{1^{n}}{6^{n}} = 
(\frac{1}{6})^{n}$

$P(\bigcap\limits_{i=1}^{6}A_{i}) = 0$

Thus, 
$P(\bigcup\limits_{i=1}^{6}A_{i}) = 6(\frac{5}{6})^{n} - 
\binom{6}{2}(\frac{4}{6})^{n} + \binom{6}{3}(\frac{3}{6})^{n} - 
\binom{6}{4}(\frac{2}{6})^{n} + \binom{6}{5}(\frac{1}{6})^{n}$

Exercise 1.49

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Exercise 1.49

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