BITSAT 2015 Paper

Total number of coins= 2n+1
Consider the following events:
E1 = Getting a coin having head on both sides from the bag. E2 = Getting a fair coin from the bag
A = Toss results in a head
Given: P (A) =

31

42

, P (E1) =

n

2n+1

and P(E2) =

n+1

2n+1

Then,
P (A) = P (E1) P (A/E1) + P (E2) P (A/E2)
⇒

31

42

=

n

2n+1

× 1 +

n+1

2n+1

×

1

2

31

42

=

n

2n+1

+

n+1

2(2n+1)

⇒

31

42

=

3n+1

2(2n+1)

⇒

31

21

=

3n+1

2n+1

n = 10