Islamic University of Gaza
Faculty of Engineering
Electrical Engineering Department
Probability and Stochastic Processes
EELE 3340
Fall semester 2011/2012
Problem Set #3 Solution
Problem(1)
Each time a modem transmits one bit, the receiving modem analyzes the signal that arrives and decides
whether the transmitted bit is 0 or 1. It makes an error with probability p, independent of whether any other
bit is received correctly.
a) If the transmission continues until the receiving modem makes its first error, what is the PMF of X, the
number of bits transmitted?
b) If p = 0.1, what is the probability that X = 10?
c) What is [𝑋 ≥ 10] ?
d) If the modem transmits 100 bits, what is the PMF of Y, the number of errors?
e) If the probability of error is p = 0.01 and the modem transmits 100 bits, what is the probability of Y = 2
errors at the receivers?
f) What is 𝑃[𝑌 ≤ 2]?
Solution :
 p (1  p ) x 1
a ) X:Geometric, PX (x)= 
0
b)
0.1(0.9) x 1
PX (x)= 
0
x=1,2,3,....
otherwise
x=1,2,3,....
otherwise
P[X=10]=0.1(0.9)9  0.0387
c) P[X  10]=1  P [ X  10]
 1  {0.1  0.1*0.9  0.1*0.92  ...  0.1*0.98 }
 1  0.1{1  0.9  0.92  ...  0.98 }  0.3874
or :
P[X  10]=0.99  0.3874
100  y
p (1  p )100 y y=0,1,2,....,100

d ) Y:Binomial, PY (y)=  y 
0
otherwise

100  2
 100 
100  2
2
98
e) PY [y=2]= 

 p (1  p )
 0.01 (1  0.01)  0.185
2 
2 
100 
1
99
f) PY [y=1]= 
 0.01 (0.99)  0.379
1 
100 
0
100
PY [y=0]= 
 0.01 (0.99)  0.366
0 
P [Y  2]  PY [y=2]  PY [y=1]  PY [y=0]=0.185+0.379  0.366  0.9206
Problem(2)
A binary source generates digits 1 and 0 randomly with probabilities 0.6 and 0.4, respectively.
a)
What is the probability that two 1s and three 0s will occur in a five-digit sequence?
b) What is the probability that at least three 1s will occur in a five-digit sequence?
Solution :
5
a ) Binomial: P(X = 2)    (0.6)2 (0.4)3  0.23
 2
b ) P(X  3 )  1  P(X  3 )
P(X  3 )  P(X  0 )  P(X  1)  P(X  2 )
5
5
5
   (0.6)0 (0.4)5    (0.6)1 (0.4)4    (0.6)2 (0.4)3  0.3174
0
1 
 2
P(X  3 )  1  P(X  3 )  1  0.3174  0.68256
Problem(3)
Suppose you know that the number of complaints coming into a phone centre averages 2.4 every ten
minutes. Assume that the number of calls follows the Poisson distribution.
a)
What is the probability that there are three or fewer calls during the next 15 minutes?
b) What is the probability that there are exactly four calls during the next ten minutes?
Solution :
 x e 
x  0,1, 2,......

a ) P X (x )  
x!
0
else

  2.4 /10  0.24 call/min
,   T
,  : average rate
  T  0.24*15  3.6
P X (x  3)  P (X  0)  P (X  1)  P (X  2)  P (X  3)
e 
 e 3.6  0.0273
x!

e
P (X  1)   1
 3.6e 3.6  0.0983
1!
3.6
e 
2 e
P (X  2)   2
  3.6 
 0.177
2!
2!

3.6
3 e
3 e
P (X  3)  
  3.6 
 0.212
3!
3!
 P X (x  3)  0.0273  0.0983  0.177  0.212  0.5146
P (X  0)   x
b)
  T  0.24*10  2.4
P (X  4)   4
2.4
e 
4 e
  2.4 
 0.125
4!
4!
Problem(4)
The random variable X has CDF
0
0.2

0.5
F X (x )  
0.6
0.6+q

1
a)
b)
c)
d)
x  2
2  x  0
0x 2
2x 3
3x 4
x 4
If 𝑃(𝑋 > 3) = 0.1 , what is the value of q?
If 𝑞 = 0.25 , Find the PMF of x, 𝑃𝑋 (𝑥)
Find 𝐸[𝑋] and 𝐸[𝑋 2 ]?
Find 𝑃 𝑋 2 − 2 > 2 ?
Solution :
a)
P (X  3)  1  P (X  3)  1  F (X  3 )  1  (0.6+q)
 0.4  q  0.1
 q  0.3
x  2
0.2
0.3
x 0

0.1
x 2
b ) P X (x )  
x 3
0.25
0.15
x 4

otherwise
0
c ) E [X ]  0.2* 2  0.3*0  0.1* 2  0.25*3  0.15* 4  1.15
E [X 2 ]  0.2* 4  0.3*0  0.1* 4  0.25*9  0.15*16  5.85
d ) Y=X 2  2
x  2  y  2
x  0  y  2
x 2 y 2
x 3  y 7
x  4  y  14
0.2
0.3

0.1
P Y (y )  
0.25
0.15

0
P (X 2  2  2)  P (Y
y 2
0.3
0.3

 0.25
y 7
0.15

y  14
0
otherwise
y  2
y 2
y  2
y 2
y 7
y  14
otherwise
 2)  P (Y  7)  P (Y  14)  0.25  0.15  0.4
Problem(5)
Consider the experiment of tossing an honest coin. Let the random variable X denote the number of tosses
required until the first head appears.
a)
Find and sketch the PMF 𝑃𝑋 (𝑥) and the CDF 𝐹𝑋 (𝑥)
b) Find 𝑃[1 < 𝑥 ≤ 4]
c)
Find 𝑃[𝑋 > 4]
Solution :
 p (1  p) x 1 x  1, 2,3,..... 0.5(0.5) x 1
a ) P X (x )  

else
0
0
x 1
0.5
0.25
x2

0.125
x 3
P X (x )  
x4
0.0625
....

..
x 1
0
0.5
1 x  2

0.75
2x 3
F X (x )  
3x 4
0.875
0.9375
4x 5

..
x  1, 2,3,..... (0.5) x

else
0
b) P(1 < X  4) = F [X = 4  ]  F [X = 1  ]  0.9375  0.5  0.4375
c) P(X  4) = 1  P (X  4)  1  F [X = 4  ]  1  0.9375  0.0625
x  1, 2,3,.....
else