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Byju's Answer
Standard XII
Mathematics
Complement of an Event
A and B take ...
Question
A and B take turns in throwing two dice, the first to throw 10 being awarded the prize, show that if A has the first throw, their chance of winning are in the ratio 12 : 11.
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Solution
There are only three possible cases, wherein the sum of the numbers obtained after throwing 2 dice is 10, i.e.
4
,
6
5
,
5
6
,
4
.
∴
P
sum of the numbers is 10
=
3
36
=
1
12
P
sum of the numbers is not 10
=
1
-
1
12
=
11
12
P
any numner other than six
=
5
6
P
A
winning
=
P
10 in first throw
+
P
10 in third throw
+
.
.
.
=
1
12
+
11
12
×
11
12
×
1
12
+
.
.
.
=
1
12
1
+
11
12
2
+
11
12
4
+
.
.
.
=
1
12
1
1
-
121
144
1+a+a
2
+a
3
+
.
.
.
=
1
1
-
a
=
1
12
×
144
23
=
12
23
P
B
winning
=
1
-
P
A
winning
=
11
23
Now
,
P
A winning
P
B winning
=
12
23
11
23
=
12
11
Hence proved.
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Q.
Two persons A and B take turns in throwing a pair of dice. The first person to through 9 from both dice will be awarded the prize. If A throws first then the probability that B wins the game is
Q.
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A and B throw alternatively a pair of dice, A wins if he throws a sum of 6 before B throws a sum of 7 and B wins if he throws sum of 7 before A throws a sum of 6. Find their respective chance of winning, If A begins.
Q.
A
and
B
play for a prize;
A
is to throw a die first, and is to win if he throws
6
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B
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