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Exercise 6
Q1.
Cr =  nr  = (n n!r )! r! ;
n
 
0! = 1 ;
1! = 1
 6
6!
6!
6  5  4!
(a)   =
=
=
=15; (b) 4! = 4  3  2  1 = 24
4! 2!
4! 2  1
 2  (6  2)! 2!
 5
5!
5!
5  4  3  2!
(c)   =
=
=
=15; (d) (0!)  (3!)=(1)  (3  2  1)=6
2! 3  2  1
 3  (5  3)! 3! 2! 3!
 10 
10!
10!
10  9  8  7  6!
(e)   =
=
=
= 210
6! 4  3  2  1
 4  (10  4)! 4! 6! 4!
n
Pr = (n n!r )! ;
Q2.
(a)
(b)
Q3
0! = 1 ;
1! = 1
P3 = (5 5!3)! = 52!! = 5  32! 2! = 15 ways
5
P3 = (6 6!3)! = 63!! = 6  5 3!4  3! = 120 ways
6
x
; ( x = 1, 2, 3, 4, 5 )
15
Properties of probability distribution are
to be tested as:
(a)
P(X = x) =
P( X  x )  0 for
x = 1, 2, 3, 4, 5
5
0 1 2 3 4 5
(ii)  P( X  x ) = + + + + + =1
15 15 15 15 15 15
x0
Since, properties, (i), (ii), are satisfied,
(i)

Note:
There are two properties of
Probability Distribution, is to be tested:
P( X  x )  0
(i)
n
(ii)
 P( X  x) = 1; for
x = 1,2, .., n
x0
If properties, (i), (ii), are satisfied,
then it represents as Probability
Distribution of discrete random variable.
It is a probability distribution of discrete random variables.
5  x2
; ( x = 1, 2, 3 )
16
5  32  4
P(X = x) =
=
<0
16
16
Since, property (i) is not satisfied,
 It does not represent as Probability Distribution of discrete random variable.
(b)
P(X = x) =
Q4.
(a)
P( X  x) = c x ; ( x = 1, 2, 3, 4 )
4
1=
 P( X  x) = c[1 + 2 + 3 + 4 ] = 10 c
;  c=
x 1

1
10
The p.d.f. of X is to be
1
P( X  x ) =
x ; where x = 1,2,3,4
10
(b) (i) P( X  3) = P(X = 1) + P(X = 2)
1
1
=
(1) +
(2)
10
10
= 0.1 + 0.2 = 0.3
Or P( X  3) = 0.3 (from table)
x
P(X=x) =
1
2
3
4
Total
1
x P(X  x)
10
0.1
0.2
0.3
0.4
1.0
0.1
0.3
0.6
1.0
-
(ii) P( X  3) = 0.6
(iii) P( X  3) = P( X  3) + P( X  4) = 0.3 + 0.4 = 0.7
(iv) P(1  X  3) = P( X  1) + P( X  2) = 0.1 + 0.2 = 0.3
Q5.
(a)
X is discrete.
3
(b)
1 =  P( X  x ) = 0.1+ 0.5 +c + 0.1
x0
= 0.7 + c ;
 c = 0.3
x
0
1
2
3
Total
P(X=x)
0.1
0.5
c = 0.3
0.1
1.0
(c)
P( X  2) = P( X  2) + P( X  3) = 0.3 + 01 = 0.4
(d)
P( X  2.5) = P( X  3) = P( X  0) + P( X  1) + P( X  2) =0.1+0.5+0.3 =0.9