problem 57

1.6.2 problem 57

The desired event can be expressed as $1 0^{22} ⋃ i = 1 A_{i}$ , where $A_{i}$ is the event that the $i$ -th molecule in my breath is shared with Caesar. We can compute the desired probability using inclusion exclusion.

Since every molecule in the universe is equally likely to be shared with Caesar, and we assume our breath samples molecules with replacement, $P (n ⋂ i = 1 A_{i}) = {(\frac{1}{1 0^{22}})}^{n}$ .

Thus,

\begin{matrix} P (⋃_{i = 1}^{1 0^{22}} A_{i}) & = \sum_{i = 1}^{1 0^{22}} {(- 1)}^{i + 1} {(\frac{1}{1 0^{22}})}^{i} = {(1 - \frac{1}{1 0^{22}})}^{1 0^{22}} \approx e^{- 1} \end{matrix}

[next] [prev] [prev-tail] [front] [up]