1. THE UNIFORM DISTRIBUTION’S EXAMPLE
Old Faithful erupts every 91 minutes. You arrive there at random and wait for
20 minutes ... what is the probability you will see it erupt?
Solution:
This is actually easy to calculate, 20 minutes out of 91 minutes is:
p = 20/91 = 0,22 (to 2 decimals)
But let's use the Uniform Distribution for practice.
To find the probability between a and a+20, find the blue area:
Area = (1/91) x (a+20 - a) = (1/91) x 20 = 20/91 = 0,22 (to 2 decimals)
So there is a 0,22 probability you will see Old Faithful erupt.
If you waited the full 91 minutes you would be sure (p=1) to have seen it erupt.
But remember this is a random thing! It might erupt the moment you arrive, or
any time in the 91 minutes.
2. THE EXPONENTIAL DISTRIBUTION’S EXAMPLE
Suppose that the amount of time one spends in a bank is exponentially
distributed with mean 10 minutes, λ = 1/10. What is the probability that a
customer will spend more than 15 minutes in the bank? What is the probability
that a customer will spend more than 15 minutes in the bank given that he is stil
in the bank after 10 minutes?
Solution:
P(X > 15) = 𝑒−15𝜆
= 𝑒−
3
2= 0.22
P(X > 15|X > 10) = P(X > 5) = 𝑒
−
1
2 = 0.604
THE GEOMETRIC DISTRUBITON’S EXAMPLE
Suppose a batter has probability 1/3 to hit the ball. What is the chance that
he misses the ball less than 3 times? The number X of balls up to the first
success is geometrically distributed with parameter 1/ 3.
Solution :
(X ≤ 3) = 1/3 + 1/3 x 2/3 + 1/3 (2/3)² = 0.7037
THE POISSON DISTRIBUTION’S EXAMPLE
A companymakeselectitricmotors. Theprobability an simple of 300
electricmotorswillcontainexactly 5 defectivemotors?
Solution:
Theavaragenumber of defective in 300 motors = 0,01*300=3
Theprobabilityof getting 5 defective is
3. P(x)=
𝑒−335
5!
= 0,10082
THE BINOMIAL DISTRIBUTION’S EXAMPLE
Suppose that you are rolling a diceeighttimes. Find the probability that the
face with two spots comes up exactly twice.
Solution:
n=8
P(x=2) = (
8
2
).(1 −
1
6
)8−2
= 28.
56
68 = 0.26
THE BERNOULLI DISTRIBUTION’S EXAMPLE
Suppose our class passed (C or better) the last exam with probability 0.75. Let
the random variableX be the probability that someone passes the exam.
Solution:
X ˜ bernoulli (0.75)
X 0 1
P(x) 0.25 0.75
E(X) = p = 0.75
Var(x) = p(1-p) = 0.75 x 0.25 = 0.1875
