Test 1 Probability and Statistics I, Friday April 3, 3.30-5.15
A (non-programmed) calculator is allowed, but not necessary.
Give short, but motivated answers to the questions.
1. Given are the following probabilities:
,
and
.
a. Compute
.
b. Are A and B independent?
2. For a master class on probability theory 6 Mathematics, 15 Computer Science and 9 Physics majors applied. Only
20 students of these 30 students can be admitted. To determine who is allowed to participate, lots are drawn (at
random).
Find the following probabilities (you can give a numerical expression as an answer: no need to use your calculator
for computing, e.g., 10!):
a. The probability that no mathematics student at all is admitted to the master class.
b. The probability that the numbers of students per major in the master class are proportional to the numbers of
students who applied (that is, the probability that the master class will consist of 4 Mathematics, 10 Computer
Science and 6 Physics majors).
3. A continuous random variable X has the following density function:
f(x) = cx for 0  x  1
and
f(x) = 0
for other values of x.
a. Show that the value of c = 2
b. Sketch the density function of X and compute E(X) and var(X)
c. Derive the density function of Y = X .
4. X and Y are independent and are both Poisson distributed, having parameters 1 and 2 respectively.
Find the standard deviation of X + Y and the correlation coefficient (X, X + Y).
5. To reduce the costs of a fast internet connection the board of a company decided that the employees have to share
one internet connection with 10 employees. The computer department of the company stated that the internet
connection is sufficiently fast for about 7 active users per connection. Not every employee will be present and not
every present employee will actively use the internet connection. Therefore the computer department supposes
that groups of 10 employees is acceptable. In this exercise we will try to check this assumption.
Based on experience of the computer department we have:
 N = “the number of present employees within an arbitrary group of 10, sharing one internet connection”.
N is homogeneously distributed on {6, 7, 8, 9, 10}: these numbers have equal probability.
 The probability that an arbitrary present employee is using the internet actively is 80%.
You can assume independence for the employees being active (at an arbitrary chosen moment) or not.
 X is the number of employees actively using internet in the arbitrary group of 10 employees.
a. Find E(N) and var(N).
b. Find P(X = 5 | N = 8), the distribution of X, given N = 8, and E(X | N = 8).
c. Compute the probability that (exactly) 8 employees, out of a group of 10, actively use internet.
------------------------------------------------The following formulas are given:
Variances: Uniform:
Exercice
Marks
Points
, Geometrical:
1
a
3
b
2
2
a
2
b
2
and Hyper geometrical:
3
a
2
4
b
2
c
3
4
5
a
2
Total
b
2
c
3
27
Mark = munber of points/ 2.7, on a
scale 0.0 - 10.0 (This mark will
determine 50% of the final mark)
Solutions:
Exercise 1
a. Because A  B  A  AB we find: P( AB)  P( A  B)  P( A)  0.8  0.35  0.45
So P( B)  P( AB)  P( AB)  0.3  0.45  0.75
Using this we have: P( A | B)  P( A  B) / P( B)  0.3 / 0.75  40%
b. A and B are not independent, because 0.3 = P(AB)  P(A)P(B) = 0.350.75 (or e.g. P(A)  P(A|B) )
Exercise 2
a. There are 20 draws without replacement from 6 Math and 24 other students: we will apply the hyper
 24   30 
geometrical formula: P(“no math student”)) =      24  ....  5 ( 0.035 %) .
11
 20   20  30
 6  15   9   30 
b. P(“4 Math, 10 CS- and 6 Physics-students”) =           ( 12 .6%) .
4
10
6
20
  
   

Exercise 3
x 1
1

1
 c 1 1
 f ( x)dx  1 , so  cxdx  c  x 2
2
2 x 0
0

x 1

1
b. E(X)=  xf ( x)dx   2 x 2dx = 2 x 3
 2.
3
x 0 3

0
a.
 c = 2.
=>
c. If Y = X , then: FY (y)= P(Y  y) = P( X  y) = P( X  y2) = FX (y2), for y ≥ 0
fY (y)= d FY (y) = 2y fX (y2)
dy
fX (y)= 2x for 0 ≤ x ≤ 1, so fY (y) = 2y fX (y2) = 4y3 for 0  y2  1 or 0  y  1
(and fY (y) = 0, elsewhere)
Exercise 4
Exercise 5
a. P( N = n ) = 1/5 for n = 6, 7, 8, 9, 10, so because of symmetry E(N) = 8
E(N 2) =  n P(N=n) = (62 + 72 + 82 + 92 + 102)/5 = 66 en var(N)= E(N2) - E(N)2 = 66 - 82 = 2
b. P(X = 5| N = 8) =
X has, given N = 8, a B(8, 0.8)-distribution: E(X | N=8) = np = 80.8 = 6.4.
c. P(X = 8) = P(X = 8 and N = 8)
+ P(X = 8 and N = 9)
+ P(X = 8 and N = 10)
= P(X = 8 | N = 8)P(N = 8) + P(X = 8 | N = 9)P(N = 9) + P(X = 8 | N = 10)P(N =10)
=
0.88 0.2
+
100.88 0.2  0.2 + 450.88  0.22 0.2
= 16.1%