Question

. I have n keys, exactly one of which opens the door. I try them one by one at random independently, removing the key from the fob and setting it aside if it doesn’t work. Let X be the number of keys I try until I open the door. Find E(X).

A. (n−1)/2

B. (n + 1)/2

C. n/2

D. n + 1/2

Answer 1

discrete random variable X = no. of keys I try until I open the door

so X can have values from 1 to n

I set aside a key if it is the wrong one .

P(X=1) = probability that we get the correct key in first try =

P(X=2) = Probability that we get the correct key in second try =

= Probability that we got wrong key in first try * Probability that we got correct key in next try

  

P(X=2) =

Similarly we get P(X=3) =

Similarly we get all the P(X=x) =

E(X) = Expected value of the random variable X

It is given by :   

So ,   

Sum of first n natural numbers is given by Sn

  

   we have ,

Option B is correct