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Sir Tanika Mukopadhyay taught us Probability at Homi Bhabha National Institute. He gave us assignments so that we can practice what we learned in form of problems. Here is solution to those problems. Its main emphasis is on following points: Analysis, Stochastic, Systems, Point, Inside, Radius, Discrete, Random, Variable, Continuous, Mixed
Typology: Exercises
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ECE 863 (^1)
Problem 1
P Y 1 P a 1 2
Problem 2
(c) P Y y y 1
y
2 2 £ = p =^2 p
ECE 863 (^2)
1 2 3 4 x
X is a discrete random variable.
Problem 4
from problem (3.4), P Y £ y = y 2 , " y £ 1
1 y
Y is a continuous random variable.
FY (y)
ECE 863 (^4)
3.14 (b) – Continued:
P 1 X 1 FX 1 - - FX^1 1 5 4 4 2 16 16
Problem 7
3.16 We have F 2
1 C 1 sin X 2
fi C =^1 2
Problem 8
e e
R R 1 2 2
s £ £ s = s - s
= -
e
R
ECE 863 (^5)
3.19 (a)
1 1 x 0 0
f x dx 1 C x 1 x dx 1
C 6
3 4 1 2
P 1 X 3 6 x 1 x dx 0. 2 4
x 2 3 x x X
F x f s ds F x 3x 2x 0 X 1 1 X 1
− ∞
Problem 10
x FX x fx s ds − ∞
x 2 2
/ (^) ds − ∞s
= α^ π
x 2
1/ (^) d s − ∞ 1 s
= π^ ^ (^) α
(^1) tan 1 s x p a
fi F x 1 tan x X 2
p a
p
ECE 863 (^7)
In other words, since a value “zero” is one of the possible values that N can have, we need to use the above form of the Geometric R.V. probability mass function
(a) P N k p 1 p j k 1
> = - j = +
=
p 1 p 1 1 1 p
1 p
k 1 i 0
i
k 1 k 1
(b) F (^) N x P N x p 1 p x 0 j 0
x (^) j
=
where x is the largest integer £ x
fi =
F x p^1 1 p + 1 1 p N^1 1 p^ x^0
x 1
(c) P N is an even numbered p 1 p j 0
2 p
(d) P N k / N m
P N k N m P N k
P N k P N m
for 1 k m
fi =
P N k 1 P N m
1 p p 1 1 p
k k 1
ECE 863 (^8)
P M k j/ M j
P M k j M j P M j P M k j P M j
In general, for a geometric random variable
P M n 1 p p i n
i n
ai 1 an 1 a n^ an^1 =^1 a
fi ≥ =
P M n 1 p - p
p 1 p
n 1
fi
P M k j - P M j
1 p 1 p
1 p
k j 1 j
Therefore, P M ≥ k + j/ M > j = P M ≥k
Problem 14
3.36 We have to show that the memory-less property:
P M ≥ k = P M ≥ k + j/ M >j
for a random variable M, leads to:
ECE 863 (^10)
3.37 (a) p O! o e^ e
(^0 ) = l^ -^ l = -
k!
e 0. k 0
9 k 15
=
Problem 16
3.40 P X > 4 < 0.9 ¤ P X £ 4 >0.
k!
e
3 / n k!
e k 0
4 k k 0
If n = 1 fi P X £ 4 @0.
fi One employee is sufficient
P X = 0 = e -^ a = e -^3 = 0.
Problem 17
( )
P error/ 1 P Y 0 / 1}
P 1 N 0 P N 1
Q 1 0.
υ = −^ =^ ≥^ υ = −
= (^) − + ≥ (^) = (^) ≥
= =
( ) ( )
P error/ 1 P Y 0 / 1
2P 1 N 0 2P N 1
1 Q 1 Q 1 0.
υ = = < υ =
= + < = < −
= − − = =
