A bag contains (2

Total coins = 2n + 1

Number of coins with both head = n

Number of coins with one head = 2n + 1 – n = n + 1

Sample Space, as n coins have head both side and n + 1 have heads only one side

Total Number of heads = 2n + n + 1 = 3n + 1

Total number of tails = n + 1

Now, Favorable outcomes = Number of heads = 3n + 1

Possible outcomes = 3n + 1 + n + 1 = 4n + 2

Let E be the event of getting a head

Given,

⇒ 31(4n + 2) = 42(3n + 1)

⇒ 124n + 62 = 126n + 42

⇒ 2n = 20

⇒ n = 10