Solutions of mid-sem exam, probability theory I,
B. math. I year, 2008-09, ISI-BC
Solution 1:
P {A ⊂ B} =
n
X
P (A ⊂ B : |B| = k) · P (|B| = k)
k=0
k
n
X
1 n X 1 k
2n k r=0 2n r
k=0
n 1 X n k
2
= n
k
4
=
k=0
3n
= n.
4
Solution 2:
Probability that there is no 1 in n trials is, (p0 + p2 )n = (1 − p1 )n .
Probability that there is no 2 in n trials is, (p0 + p1 )n = (1 − p2 )n .
Therefore, probability that outcomes 1 and 2 both occur atleast once is,
1 − (1 − p1 )n − (1 − p2 )n + pn0 . Because we were excluding the case 2 times where
only 0 appears in n trials.
Solution 3:
We will prove it by mathematical induction. Assume it holds for (n − 1) trials. In
the n th trial, it could be(i) Either, we have odd number of success, we will need the last trial to be a success
to have even number of success.
(ii) OR, we have even number of success, then we need the last trial to be failure.
Pn = (1 − Pn−1 )p + Pn−1 (1 − p)
1 + (1 − 2p)n−1
1 + (1 − 2p)n−1
p+
(1 − p)
= 1−
2
2
1 + (1 − 2p)n−1
1 + (1 − 2p)n−1
1 + (1 − 2p)n−1
=p−p
+
−p
2
2
2
1
1
n−1
n−1
= − p(1 − 2p)
+ (1 − 2p)
2
2
1
n−1 1
( − p)
= + (1 − 2p)
2
2
1 + (1 − 2p)n
=
.
2
1
2
Solution 4:
E
n
1 X 1
n x
=
p (1 − p)n−x
x+1
x
+
1
x
x=0
=
n
X
n!
1
px (1 − p)n−x
x
+
1
x!(n
−
x)!
x=0
n
X
=
1
(n + 1)!
px+1 (1 − p)n−x
(n
+
1)p
(x
+
1)!(n
−
x)!
x=0
=
n
X
(n + 1)!
1
px+1 (1 − p)n−x
(n + 1)p x=0 (x + 1)!(n − x)!
=
1 X
m!
py (1 − p)m−y
mp y=1 y!(m − y)!
=
m!
1
1 X
py (1 − p)m−y −
(1 − p)m
mp y=0 y!(m − y)!
mp
m
[assuming x + 1 = y and n + 1 = m]
m
(1 − p)m
(p + (1 − p))m
−
mp
mp
1 − (1 − p)m
=
mp
1 − (1 − p)n+1
. [by replacing m = n + 1]
=
(n + 1)p
=
Solution 5:
λE[(X + 1)n−1 ]
=λ
∞
X
(x + 1)n−1
x=0
=
=
=
∞
X
(x + 1)n
x=0
∞
X
e−λ λx+1
(x + 1)!
(x + 1)n
x+1=1
∞
−λ x
X
λ
ne
x
x=0
n
= E[X ].
x!
e−λ λx
x!
e−λ λx+1
(x + 1)!
3
Now, if X is a Poisson random variable with parameter λ, then E[X] = λ. And,
by the above formula,
E[X 2 ] = λE[(X + 1)]
= λE[X] + λ
= λ2 + λ.
Similarly, by applying the above formula,
E[X 3 ] = λE[(X + 1)2 ]
= λE[X 2 + 2X + 1]
= λE[X 2 ] + 2λE[X] + λ
= λ3 + λ2 + 2λ2 + λ
= λ3 + 3λ2 + λ.
Solution 6:
Let there are i (1 ≤ i ≤ n) elements in the choosen subset. So,
n
P (X = i) =
i
2n − 1
.
Therefore,
E[X] =
n
X
i
n
i
2n − 1
n n X n−1
= n
2 − 1 i=1 i − 1
n
= n
2n−1
2 −1
n
=
n−1 .
2 − 12
i=0
V ar(X) = E[X 2 ] − E 2 [X]
= E[X(X − 1) + X] − E 2 [X]
= E[X(X − 1)] + E[X] − E 2 [X].
Now,
E[X(X − 1)] =
n
X
i(i − 1)
n
i
2n − 1
n n(n − 1) X n − 2
= n
2 − 1 i=2 i − 2
i=0
=
n(n − 1) n−2
2
.
2n − 1
4
Therefore,
V ar(X) = E[X(X − 1)] + E[X] − E 2 [X]
=
=
=
=
=
=
n(n − 1) n−2
n
n2
n−1
22n−2
2
2
+
−
2n − 1
2n − 1
(2n − 1)2
(n2 − n)2n−2 (2n − 1) + n2n−1 (2n − 1) − n2 22n−2
(2n − 1)2
2
2n−2
n−2
(n − n)(2
−2
) + n22n−1 − n2n−1 − n2 22n−2
(2n − 1)2
2 2n−2
2n−2
n 2
− n2
− n2 2n−2 + n2n−2 + n22n−1 − n2n−1 − n2 22n−2
(2n − 1)2
2n−2
2 n−2
n−2
n2
−n 2
− n2
[since, n2n−1 = 2n2n−2 and n22n−1 = 2n22n−2 ]
n
2
(2 − 1)
n22n−2 − n(n + 1)2n−2
.
(2n − 1)2
Solution 7:
(a) From the given conditions P {X = 2} = 21 · 13 .
1
1
Similarly P {X = i} = 12 · 23 · 34 · · · i−1
i · i+1 = i(i+1) .
P∞
1
Therefore P {X > i} = k=i+1 k(k+1)
, where i ≥ 1.
(b)
P {X < ∞}c = lim P (X ≥ i)
i→∞
∞
X
1
k(k + 1)
k=i
N X
1
1
= lim lim
−
i→∞ N →∞
k k+1
= lim
i→∞
k=i
1
= lim
i→∞ i
= 0.
Therefore P {X < ∞} = 1.
(c)
E[X] =
∞
X
i=1
∞
i·
X 1
1
=
= ∞.
i(i + 1)
i+1
i=1