264 16. Answers to Exercises
In the formula of Alice, the rounding plays less and less of a role, so
⌈
en/^2 −^1⌉
∼en/^2 −^1 =(0. 367 ...)·(1. 648 ...)n,and so the ratio between Alice’s numbers and the corresponding Fibonacci
numbers is
en/^2 −^1
Fn≈
(0. 367 ...)·(1. 648 ...)n
0. 447 ...=(0. 822 ...)·(1. 018 ...)n.Since the base of the exponential is larger than 1, this tends to infinity as
ngrows.
5 Combinatorial Probability
5.1 Events and Probabilities
5.1.1. The union of two eventsAandBcorresponds to “AorB”, i.e., at
least one ofAorBoccurs.
5.1.2. It is the sum of some of the probabilities of outcomes, and even if
we add up all of the probabilities, we get just 1.
5.1.3. P(E)=^12 ,P(T)=^13.
5.1.4. The same probabilitiesP(s) are added up on both sides.
5.1.5. Every probabilityP(s) withs∈A∩Bis added twice to both sides;
every probabilityP(s) withs∈A∪Bbuts/∈A∩Bis added once to both
sides.
5.2 Independent Repetition of an Experiment
5.2.1. The pairs (E, T),(O, T),(L, T) are independent. The pair (E, O)is
exclusive. Neither the pair (E, L) nor the pair (O, L) is independent.
5.2.2. P(∅∩A)=P(∅)=0=P(∅)P(A). The setSalso has this property:
P(S∩A)=P(A)=P(S)P(A).
5.2.3. P(A)=|S|
n− 1
|S|n =1
|S|,P(B)=|S|n−^1
|S|n =1
|S|,P(A∩B)=|S|n−^2
|S|n =
1
|S|^2 =P(A)P(B).5.2.4. The probability that your mother has the same birthday as you
is 1/365 (here we assume that birthdays are distributed evenly among all
numbers of the year, and we ignore leap years). There are (roughly) 7 billion
people in the word. This means that one could expect that 7· 109 /365 (about
20 million) people have the same birthday as their mother. The events that
your birthday coincides with that of your mother, father, or spouse are
independent, so the probability that for a given person, all three were born