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Byju's Answer
Standard XII
Mathematics
Conditional Probability
solve Q14....
Question
solve
Q14. A , B and C throw a die alternately till one of them gets a '6' and wins the game. Find their respective probability of wining. If A starts the game.
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Solution
Dear student
Let
E
be
the
event
of
getting
a
six
in
a
single
throw
of
an
unbiased
die
.
Then
,
P
(
E
)
=
1
6
and
P
E
¯
=
1
-
1
6
=
5
6
A
wins
if
he
gets
a
six
in
1
st
or
4
th
or
7
th
.
.
.
throw
.
His
probability
of
getting
a
'
six
'
in
first
throw
=
P
(
E
)
=
1
6
A
will
get
fourth
throw
if
he
fails
in
first
,
B
fails
in
second
and
C
fails
in
third
throw
.
So
,
Probability
of
winning
of
A
in
fourth
throw
=
P
E
¯
∩
E
¯
∩
E
¯
∩
E
=
P
E
¯
P
E
¯
P
E
¯
P
E
=
5
6
3
×
1
6
Similarly
,
the
probability
of
winning
of
A
in
7
th
throw
=
5
6
6
×
1
6
and
so
on
.
.
.
Hence
probability
of
winning
of
A
=
1
6
+
5
6
3
×
1
6
+
5
6
6
×
1
6
+
.
.
.
=
1
6
1
-
5
6
3
Sum
of
infinte
G
.
P
=
a
1
-
r
and
here
a
=
1
6
and
r
=
5
6
3
=
36
91
B
wins
if
he
gets
'
six
'
in
2
nd
or
5
th
or
8
th
or
.
.
.
throw
So
,
probability
of
winning
of
B
=
5
6
×
1
6
+
5
6
4
×
1
6
+
5
6
7
×
1
6
=
5
6
×
1
6
1
-
5
6
3
Sum
of
infinte
G
.
P
=
a
1
-
r
and
here
a
=
5
6
×
1
6
and
r
=
5
6
3
=
30
91
Hence
probability
of
winning
of
C
=
1
-
36
91
+
30
91
=
25
91
Regards
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