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Byju's Answer
Standard VIII
Mathematics
Experiment and Outcomes
a, b, c throw...
Question
a, b, c throw a die alternatively by till one of them gets a 6 and wins a game.find the respective probabilities of winning if a starts first.
will we find probabilities of winning of all three i.e. a b and c .
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Solution
Let
E
be
the
event
of
getting
a
six
in
a
single
throw
of
the
die
.
P
E
=
1
6
⇒
P
E
=
1
-
1
6
=
5
6
A
wins
if
throws
six
in
1
st
or
4
th
or
7
th
or
.
.
.
throws
.
Also
,
B
does
not
throw
a
six
in
2
nd
or
5
th
or
8
th
.
.
.
throws
and
C
does
not
throw
a
six
in
3
rd
or
6
th
or
9
th
.
.
.
throws
.
P
A
wins
=
P
E
∪
E
∩
E
∩
E
∩
E
∪
E
∩
E
∩
E
∩
E
∩
E
∩
E
∩
E
∪
.
.
.
=
P
E
+
P
E
∩
E
∩
E
∩
E
+
P
E
∩
E
∩
E
∩
E
∩
E
∩
E
∩
E
+
.
.
.
=
P
E
+
P
E
P
E
P
E
P
E
+
P
E
P
E
P
E
P
E
P
E
P
E
P
E
+
.
.
.
=
1
6
+
5
6
3
1
6
+
5
6
6
1
6
+
.
.
.
=
1
6
1
-
5
6
3
=
36
91
B
wins
if
throws
six
in
2
nd
or
5
th
or
8
th
or
.
.
.
throws
.
Also
,
A
does
not
throw
a
six
in
1
st
or
4
th
or
7
th
.
.
.
throws
and
C
does
not
throw
a
six
in
3
rd
or
6
th
or
9
th
.
.
.
throws
.
P
B
wins
=
P
E
∩
E
∪
E
∩
E
∩
E
∩
E
∩
E
∪
E
∩
E
∩
E
∩
E
∩
E
∩
E
∩
E
∩
E
∪
.
.
.
=
P
E
∩
E
+
P
E
∩
E
∩
E
∩
E
∩
E
+
P
E
∩
E
∩
E
∩
E
∩
E
∩
E
∩
E
∩
E
+
.
.
.
=
P
E
P
E
+
P
E
P
E
P
E
P
E
P
E
+
P
E
P
E
P
E
P
E
P
E
P
E
P
E
P
E
+
.
.
.
=
5
6
×
1
6
+
5
6
4
1
6
+
5
6
7
1
6
+
.
.
.
=
5
36
1
-
5
6
3
=
30
91
Now
,
P
A
wins
+
P
B
wins
+
P
C
wins
=
1
⇒
36
91
+
30
91
+
P
C
wins
=
1
⇒
P
C
wins
=
1
-
36
91
-
30
91
=
25
91
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