Q. 23

# A bag contains (2

Answer :

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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