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BITSAT 2015 Solved Question Paper
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© examsnet.com
Question : 88
Total: 150
A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is 31/42, then n is equal to
10
11
12
13
Validate
Solution:
Total number of coins= 2n+1
Consider the following events:
E
1
= Getting a coin having head on both sides from the bag.
E
2
= Getting a fair coin from the bag
A = Toss results in a head
Given: P (A) =
31
42
, P
(
E
1
)
=
n
2
n
+
1
and
P
(
E
2
)
=
n
+
1
2
n
+
1
Then,
P (A) = P
(
E
1
)
P (A/
E
1
) + P
(
E
2
)
P (A/
E
2
)
⇒
31
42
=
n
2
n
+
1
× 1 +
n
+
1
2
n
+
1
×
1
2
31
42
=
n
2
n
+
1
+
n
+
1
2
(
2
n
+
1
)
⇒
31
42
=
3
n
+
1
2
(
2
n
+
1
)
⇒
31
21
=
3
n
+
1
2
n
+
1
n = 10
© examsnet.com
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