MathDB
Problems
Contests
National and Regional Contests
India Contests
India LIMIT
2019 LIMIT
2019 LIMIT Category C
Problem 11
Problem 11
Part of
2019 LIMIT Category C
Problems
(2)
alternating series
Source: LIMIT 2019 CCS1 P11
4/28/2021
Let
x
=
1
1
⋅
2
−
1
2
⋅
3
+
1
3
⋅
4
−
…
x=\frac1{1\cdot2}-\frac1{2\cdot3}+\frac1{3\cdot4}-\ldots
x
=
1
⋅
2
1
−
2
⋅
3
1
+
3
⋅
4
1
−
…
Then
e
x
+
1
e^{x+1}
e
x
+
1
is
algebra
find conditional distribution
Source: LIMIT 2019 CCS2 P11
4/28/2021
Let
X
1
,
X
2
,
X
3
X_1,X_2,X_3
X
1
,
X
2
,
X
3
be
exp
(
1
)
\exp(1)
exp
(
1
)
. Find the conditional distribution of
X
1
∣
X
1
+
X
2
+
X
3
=
k
X_1|X_1+X_2+X_3=k
X
1
∣
X
1
+
X
2
+
X
3
=
k
.
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
Uniform
(
0
,
k
)
<span class='latex-bold'>(A)</span>~\operatorname{Uniform}(0,k)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
Uniform
(
0
,
k
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
Uniform
(
0
,
k
3
)
<span class='latex-bold'>(B)</span>~\operatorname{Uniform}\left(0,\frac k3\right)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
Uniform
(
0
,
3
k
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
Uniform
(
0
,
2
k
3
)
<span class='latex-bold'>(C)</span>~\operatorname{Uniform}\left(0,\frac{2k}3\right)
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
Uniform
(
0
,
3
2
k
)
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
None of the above
<span class='latex-bold'>(D)</span>~\text{None of the above}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
None of the above
probability