Test information P&S I: test Friday April 3, 3.30 MDI-1
Topics: Introduction, Combinatorics, Conditional Probability, Distributions and Joint discrete distributions.
Review and prepare by making the following exercises as a trial test: Introduction: 6, Conditional
Probability: 7 , Distributions: 10, 16, 24.b and Joint discrete distributions: 6.a,b.
A second trial Test (find an overview of the formulas you have to know after this trial test):_____________
The use of a calculator is allowed. Tables are given if necessary.
The following formulas are given:
Variances: Uniform:
, Geometrical:
and Hyper geometrical:
1. State the definition of the correlation coefficient and compute
2.
3.
4.
5.
if X and Y are independent and Poisson
distributed random variables, both having parameter µ = 2.
A person claims to clairvoyant (can see things, that “normal” people can not see). To test his claim he is shown
ten closed boxes. In each of these boxes there is small bottle that is filled with either water or oil (with equal
probability for each box). The person has to guess (“see”) whether each of the 10 boxes contains a bottle with
water or a bottle with oil. X is the number of correct answers he gives.
Determine P(X ≥ 9), E(X) and var(X), supposing he is not clairvoyant at all.
A gynaecologist found out that about 60% of the women who are pregnant according to a “home-test” for
pregnancy are not pregnant at all. The producer of the test claims that the test is positive for 100% of all pregnant
women and will be negative for 90% of non-pregnant women.
Assume that both claims (of gynaecologist and producer) are correct and find the probability p that a user of the
test is pregnant.
The random variable X has a probability density function given by the formula:
f(x) = c x2 for 0 ≤ x ≤ 1 and f(x) = 0 for other values of x. (c is a unknown constant).
a. Show that c = 3.
b. Compute E(X) and var(X).
c. Find the density function of Y = X 2
(The wine tasting problem) Somebody claims to be a connoisseur of wine. To test his expertise he gets 6 cards
with the names of 6 types of red wine. He is presented 6 glasses of these wines and, after tasting them, he has to
put one card at each glass. Suppose he is not an expert on wines at all. What is the probability he will guess all
wines right?
a. What is the probability he will guess all wines right?
For i = 1, 2, ..., 6 we define Xi = 1 if he gives the i-th glass the correct name and Xi = 0 otherwise.
So the number of correct named wine types is S = X1 +....+ X6
b. Find E(X1) and var(X1)
c. Are X1,...., X6 independent? Motivate your answer.
d. Compute E(S), the expected number of correctly named wine types.
e. Compute cov(X1, X2) and var(S).
---------------------------------------------------------------------------------
Overview: Definitions:
E(X) =…………………………........................(discrete) E(X) =……………………...........(continuous)
var(X) = …………………………………………. cov(X, Y) =………………………………………..
Formulas
Eg(X) =………………………….........(discrete). Eg(X) =………………………...................(continuous)
Eg(X,Y) =……………………….........(discrete).
Var(X) = …..E(X2 ) …………………
cov(X, Y) = ……E(XY)………………..
E(aX + b) =….E(X)……………….
var(aX + b) =……var(X)………………
E(X + Y) =…..E(X)………………..
var(X + Y) =….. var(X)……………..
E(X - Y) = …..E(X)………………….
var(X - Y) =…….var(X)…………….
E(X1+…+Xn) = E(X1)…………………
var(  Xi )  …………………………..
cov(aX, Y) =……cov(X, Y)…………..
cov(X, X) =……………………..
n
i 1
cov(X, Y + Z) = ….cov(X, Y)……………………………………….
Independence
X en Y are independent  P(X = x and Y = y) = …………………………………….
(discrete)
=> E(XY)= ……..., cov(X, Y) = ……….., var (X ± Y) = …………………….
n
X1,…,Xn are independent and all have variance σ2 => var(  Xi )  …………………………………
i 1
Common Distributions:
Name + parameter(s)
Binomial
B(n, p)
Poisson
µ
Geometrical
p
Hypergeometrical
R, N, n
Uniform
U(a,b)
Exponential
E(λ)
Standardnormal
N(0,1)
Normaal
N(µ, σ2)
Probability (density) function
E(X)
Var(X)
P(X = k) =…………………………..
……
.............
P(X = k) =
……
.............
1 p
P(X = k) =………………………….
p2
……
……
P(X = k)= …………………………


N n
nR
1 R
N
N N 1
f(x) =………………..for a ≤ x ≤ b
……
..............
f(x) = ……………………..
……
..............
( x )  …………………………
……
................
……
........................
f(x) =
Distribution X
Hyper geometrical
R, N, n
Relations
between
Binomial
distributions B(n, p)
Normal
X ~ N(µ, σ2)
1
2 2
x 
 1 

2  
e
2
(approximating) distribution
Rule of thumb
Binomial B(…...,……..)
N > ……….
Poisson, µ = …..
Normal
Y = aX +b ~ N(……… ,…………)
np(1-p) <……..
Solutions for the trial test 1
cov( X , Y )
1)  ( X , Y ) 
. We only use that var(X) = var(Y) and that X and Y are independent
 XY
cov( X , X  Y ) cov( X , X )  cov( X , Y ) var( X )  0
1
(X , X  Y ) 



2
 X  X Y
 X var( X )  var(Y )
 X 2  2X
2) X ~ B(10, 1/2),
 10   10    1 10
11
 1.07% E ( X )  np  10  12  5
so P( X  9)  P( X  9)  P ( X  10)           
9
10
2
1024


    
var(
X
)

np
(
1

p
)

2
.
5
and
3) Define A = “User of selftest is really pregnant” and T = “Test is positive (indicates pregnancy)”.
Given: P (T | A)  1 , P(T | A)  0.10 en P( A | T )  0.60
Question p  P( A) = ? . Solution:
=> 0.06  0.54 p  0.10  0.1 p , or p 

1

0
0.04
0.64
 6.25%
4) a.  f ( x)dx  1   c x2 dx  13 c x3 |xx 10  13 c  1
=>
c3
1
b. E ( X )   x  3 x2 dx  34 x4 |xx 10 
3
4
0
1
E ( X 2)   x2  3 x2 dx  53 x54 |xx 10 

3
5
var( X )  E( X 2)  EX 2  53  169  803
0
y 0
c. F Y ( y)  P( X 2  y)  P( y  X  y )  F X ( y )  F X ( y )
1
f Y ( y )  dyd F Y ( y ) 
[ f X ( y )  f X ( y )] .
2 y
Now we use that f X ( x)  3 x2 for 0  x  1 , so:
1
[3 y  0]  32 y , for 0  y  1
f Y ( y) 
2 y
(and f Y ( y)  0 for all other values of y)
1
1

6! 720
b. X 1 ~ B(1, 1/6), so E ( X 1)  16 and var( X 1)  365
c. No: if X 1 =…= X 5 =1, then X 6 =1, (or, e.g., 16  P( X 1  1)  P( X 1  1 | X 2  1)  15 )
5) a. P("all names correct") 
6
6
i 1
i 1
d. E ( S )  E ( X i )   E ( X i )  6 E ( X 1)  6  16  1
e. E ( X 1 X 2)    x1x 2P( X 1  x1 and X 2  x 2)  1  1  P( X 2  1 | X 1  1) P( X 1  1)  15  16
1
cov( X 1 , X 2)  E ( X 1 X 2)  E ( X 1) E ( X 2)  15  16  16  16  180
6
6
i 1
i 1
1
var( S )  var(  X i )   var( X i )    cov( X i , X j )  6  365  30  180
1
1
3
2
4
3
3
a
1
Mark = (#points +2)/2.5
b
2
i j
4
c
2
d
1
e
2
a
1
5
b c
2 2
Total
23

Overview: Definitions: E(X) =  xP( X  x ) or E(X) =  xf ( x )dx (continuous)

x
var(X) = E(X – EX)
2
cov(X, Y) = E(X – EX)(X – EY) Formulas Eg(X) =
 g( x )P( X  x ) (discrete).
x

Eg(X,Y) =  g( x , y )P( X  x  Y  y ) (discrete).
Eg(X) =  g ( x ) f ( x )dx (continuous)

x y
Var(X) = E(X2 ) – (EX) 2
E(aX + b) = aE(X) + b
E(X + Y) = E(X)+ E(Y)
E(X - Y) = E(X) – E(Y)
cov(X, Y) = E(XY) – E(X)E(Y)
var(aX + b) = a2 var(X)
var(X + Y) = var(X)+ var(Y) + 2cov(X,Y)
var(X - Y) = var(X)+ var(Y) - 2cov(X,Y)
E(X1+…+Xn) = E(X1) +…+E(Xn)
var(  Xi )   var X i +   cov( X i , X j )
n
n
i 1
i 1
i j
cov(aX, Y) = a cov(X, Y)
cov(X, X) = var(X)
cov(X, Y + Z) = cov(X, Y) + cov(X, Z)
Independence: X en Y are independent  P(X = x and Y = y) = P(X = x)×P(Y = y), (discrete)
=> E(XY)= E(X)E(Y), cov(X, Y) = 0, var (X ± Y) = var(X) + var(Y)
n
X1,…,Xn are independent and all have variance σ2 => var(  Xi )  n σ2
i 1
Name + parameter(s)
Binomial
B(n, p)
Poisson
µ
Probability (density) function
n
P(X= k) =   p k 1 p nk , k = 0,...,n
k 
P(X= k) =
 k 
e , k= 0,1,…
k!
Geometrical
p
P(X= k) = 1 p k 1 p , k = 1,2,…
Hypergeometrical
R, N, n
 R  N  R 
 

 k  n  k 
P(X= k) =
, k = 0, 1,..,n
N
 
n
Uniform
U(a,b)
Exponential
E(λ)
Standardnormal
N(0,1)
Normaal
N(µ, σ2)
Relations
between
distributions
f(x) = 1 , for a ≤ x ≤ b
b a
f(x) = e  x , for x ≥ 0
E(X)
Var(X)
np
np(1-p)
µ
µ
1
p
1 p
np
(p=R/N)
np(1-p)×
a b
2
1

( b  a )2
12
1
2
0
1
µ
σ2
p2
N n
N 1
1 2
 x
( x )  1 e 2
2
f(x) =
1
2  2
Distribution X
Hyper geometrical R, N, n
Binomial B(n, p)
Normal X ~ N(µ, σ2)
 x  
1

2  
e
2
(approximating) distribution
Binomial B(n, R/N)
Poisson, µ = np
Normal Y = aX +b ~ N(aµ+b, a2σ2)
Rule of thumb
N > 5n2
np(1-p) < 5