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Question 1
Four nickels and six dimes are tossed, and
the total number N of heads is observed.
If N = 4, what is the conditional probability
that exactly two of the nickels were heads?
Question 1A is the event that N = 4.
B is the event that exactly two of the nickels
were heads.
B
A
Question 1Let X1 be the number of heads of the nickels and X2 thenumber of heads of the dimes. Then, N = X1 + X2
P(X1 = 2 | N = 4) = P(X1 = 2,X1 + X2 = 4) / P(X1 + X2 = 4)= P(X1 = 2, X2 = 2) / P(X1 + X2 = 4)= P(X1 = 2)P(X2 = 2) / P(X1 + X2 = 4)
(assuming the probabilities of heads are 0.5)
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Question 2
A dice is rolled and the number N on the uppermost face is recorded. From a jar containing 10 tags numbered 1, 2, ..., 10 we then select N tags at random without replacement.
Let X be the smallest number on the drawn tags.
Determine P(X = 2).
Question 2
90
102
10
18
11 nn
C
CCnNXP
n
n
10 to1 from tags#
10 to3 from tags1-n and 2 is tag1#
10 to1 from tags#
2number smallest with 10 to1 from tags#2
that,Note
226
1
n
n
nnNXP
nNXPnNPXPn
Question 2
2204.090
10
6
1
2
22
6
1
6
1
6
1
n
n
n
nn
nNXPnNP
nNXPnNPXP
Question 3
Let X be a Poisson random variable with
parameter, .
Find the conditional mean of X, given that X
is odd.
Question 3(define that 0 is even) Let pk = P(X = k).
0k
oddXkXkPoddXXE
4321
4321
34
2321X
2
4321
4321
432sG
p.g.f.,by Similarly
212111
,p.g.f.by that Note
ppppG
ppppG
spspspp
eGevenXPoddXP
X
X
X
Question 3
21
!
,,0,1,2,i 1,2ik is that odd, isk if
0
,0,1,2,i 2i,k is that even, isk if
212113
have, We.sG that Note
2
231
1X
e
ke
oddXP
kXPoddXkXP
oddXkXP
eGGpp
e
k
XX
s
Question 3
coth21
21
321
1
121221
1
21
!1212
1212
Thus,
2
2
312
02
02
12
0
e
e
ppe
iXPie
e
iei
oddXiXPioddXXE
i
i
i
i
Note: coth (.) denotes hyperbolic cotangent.
Question 3 - OptionalAn alternative solution without using p.g.f.:
!6!4!21
1
.2
!7
7
!5
5
!3
31
1
.2
!7.7
!5.5
!3.3.
1
2
21
!1212
1212
642
2
642
2
753
2
02
12
0
e
e
e
e
eeeee
e
iei
oddXiXPioddXXE
i
i
i
Question 3 - Optional
Note: cosh (.) denotes hyperbolic cosine.
coth1
1
!6!4!21cosh2
!4!3!21
!4!3!21
,!6!4!2
1cosh,
2
2
642
432
432
642
e
eoddXxXE
ee
e
e
Or
But
Question 4
Suppose that N has density function:
P(N =n) = (1-p) n-1p for n = 1, 2,…
where p in (0, 1) is a parameter.
This defines the geometric distribution with
parameter p.
Question 4a
Show that: G(s) = sp/[1 - s(1 - p)] for s < 1/(1 - p).
Proved11
111
1
1
1
0
ps
sp
pssppps
rNPssEsG
r
r
r
rr
r
rN
Question 4bShow that: E(N) = 1/p
Proved
1
11
11
11
11
111
111
1
2
1
2
1
2
1
pps
p
ps
spppspp
ps
spppsp
ps
sp
ds
dGNE
s
s
s
s
Question 4cShow that: var(N) = (1 - p)/p2
Proved
11121112
1var
12
11
12
111
2
1111
2222
2
2
1
3
1
3
1
2
p
p
p
pp
ppp
p
NENENNEN
p
p
ps
pp
pps
p
ps
p
ds
dGNNE
s
s
s
Question 4dShow that: P(N is even) = (1 - p)/(2 - p).
Proved2
1
2
12
2
1
2
2
2
1
111
2
111
2
1
p
p
p
p
p
pp
p
pGevenNP
