## Probability and Computing: Chapter 1 Exercises (Cont. 1)

Exercise 1.8:

Let $E_{1},E_{2},E_{3}$ be the event that the number is divisible by 4,6, and 9; respectively.

It’s easy to calculate:

$Pr(E_{1}) = 0.25;Pr(E_{2}) = 0.166666;Pr(E_{3}) = 0.111111$

$Pr(E_{1}\cap E_{2}) = 0.083333;Pr(E_{2}\cap E_{3}) = 0.055555;Pr(E_{1}\cap E_{3}) = 0.027777$

$Pr(E_{1}\cap E_{2}\cap E_{3}) = 0.027777$

So the probability that the number chosen is divisible by one or more of 4, 6, and 9 is:

$Pr(E_{1}\cup E_{2}\cup E_{3}) = \sum_{i = 1}^{3} Pr(E_{i}) -\sum_{i

Exercise 1.10:

Let $E_{1}$ be the event that the chosen coin is fair, $E_{2}$ be the event it is two-headed; and let B is the event that the flip is head.

We have $Pr(E_{1}) = Pr(E_{2}) = \frac{1}{2};Pr(B\mid E_{1}) = \frac{1}{2}; Pr(B \mid E_{2}) = 1.$

$Pr(E_{2}\mid B) = \frac{Pr(B \mid E_{2})Pr(E_{2})}{\sum_{i = 1}^{2}Pr(B\mid E_{i})Pr(E_{i})} = \frac{2}{3}.$