Q23 of 29 Page 1

A bag contains (2n +1) coins. It is known that ‘n’ of these coins have a head on both its sides whereas the rest of the coins are fair. A coin is picked up at random from the bag and is tossed. If the probability that the toss results in a head is find the value of ‘n’.

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


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