Answer:
1) Probability that Peter wins = PeW = 48.53%, while the probability that Paul wins = PaW= 1-PeW = 51.47% (Peter si favoured)
2) The bet will be pretty fair at n=16 or 17
Step-by-step explanation:
Since each car number plate is independent of the others, then the random variable X=number of cars that have the same last two digits follows a binomial distribution. Thus the probability distribution is
P(X) = n!/((n-x)!*x!)*p^n*(1-p)^(n-x)
where
p= probability that a car has the same last 2 digits = {00, 11, ...,99}/{00, 01, 02, ........ 98, 99} = 10/100 = 1/10
n= number of cars = 16
Then the probability that Peter wins PeW is:
PeW=P(X≥2) = 1- P(X<2) = 1- F(X=2) = 1 - 0.5147 = 0.4853 = 48.53%
therefore
Probability that Peter wins = PeW = 48.53%
Probability that Paul wins = PaW= 1-PeW = 51.47% (Peter si favoured)
for n=17 , PeW = 0.518 , PaW= 0.482
for n=15 , PeW = 0.450 , PaW= 0.548
then the bet will be pretty fair at n=16 or 17
