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A fair coin is thrown large number of times say n and m denotes the number of times it shows up heads. As n the fraction m n is also equal to

Options

  1. AThe probability of a six digit number N whose six digits are 1,2,3,4,5,6 written as random order is divisible
  2. BThe chance that the product of the outcomes of three rolls of a fair dice is a prime number
  3. CLim _ n ( n ^ n +1 e ( n +1)^ n - n )
  4. DLim _ n _ i =1 ^ n 1 n [ 4 i n ] where [ x ] denotes the greatest integer less than or equal to x

Correct answer

C. Lim _ n ( n ^ n +1 e ( n +1)^ n - n )

Step-by-step solution

Clearly as n the fraction m n is equal to 1 2 . (A) We have n ( S )=6 ! Note that 1+2+3+4+5+6=21 divisible by 3 For divisible by 6, N must be even, so n(A)=3 5 ! Hence P(A)= 3 5! 6! = 1 2 Ans. (B) We have n ( S )=216 Product is prime only when two rolls are 1 and the third is prime are 2,3,5 . Favourable cases are 211 (3); 311 (3) and 511 (3) i.e. 9 cases Hence P(A)= 9 216 = 1 24 Ans. (C) aligned & Lim _ n n ( e (1+ 1 n )^n -1 )= Lim _ n -n (1- e (1+ 1 n )^n )= Lim _ n -n ( (1+ 1 n )^n-e (1+ 1 n )^n ) ; Put n= 1 y

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