problem 49

1.5.1 problem 49

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

By inclusion-exclusion, $P (6 ⋃ i = 1 A_{i}) = \sum_{i = 1}^{6} P (A_{i}) - \sum_{1 \leq i < j \leq 6} P (A_{i} \cap A_{j}) + \sum_{1 \leq i < j < k \leq 6} P (A_{i} \cap A_{j} \cap A_{k}) - \dots - P (6 ⋂ i = 1 A_{i}) .$

Now,

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

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

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

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

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

$P (6 ⋂ i = 1 A_{i}) = 0$

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

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