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and its probability is:
NN
N
--
-
1
N
N
-
-
1
1
NN
N
-
w
w
w
(
)
=
p
3
10
1
◊
◊
1
The probability of drawing a black ball at the third trial is them:
()
=
(
)
+
(
)
+
(
)
+
(
)
p
1
p
100
p
101
p
110
p
111
3
3
3
3
3
We wish to know the probability of drawing a white ball at the
n
th trial.
We shall denote this probability now by
p
(
n
), and that of drawing a black
ball
q
(
n
) = 1 -
p
(
n
).
By iteratively approximating [through Eq. (35)] trial tails of length
in
as
being path independent [
p
i
(
j
) =
p
1
(
j
)] one obtains a first-order approxima-
tion for a recursion in
p
(
n
):
m
N
qn m
()
=-
(
)
+
(
)
pn
pn m
-
(36)
or for
m
=
n
- 1 (good for
p
(1) ª 1, and
n
/
N
<< 1):
n
N
-
1
()
=
()
+
()
pn
p
1
q
1
(37)
and for
m
= 1 (good for
p
(1) ª 1):
1
()
=-
(
)
+
(
)
pn
pn
1
N
qn
-
1
(38)
A second approximation changes the above expression to
pn
()
=-
pn
(
1
)
+
q
qn
(
-
1
)
(39)
where q=q(
N
,
N
w
) is a constant for all trials. With this we have
()
--
(
)
==-
(
)
pn
pn
1
Dq
p
1
p
(40)
which, in the limit for
D
D
p
n
dp
dn
lim
D
=
n
Æ
0
gives
dp
dn
=-
()
(
)
q 1
pn
with the solution
()
=- -
(
)
-
pn
11
0
p e
q
n
(41)
This, in turn, is an approximation for
p
ª 1 of
p
()
=
0
pn
(42)
(
)
-
p
+-
1
p
e
q
n
0
0
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