Question

Each student that sees the notes has a1/6 chance of taking a copy of them, independent of any other student, and the Professors will stop the experiment either when someone picks up and takes a copy of the notes or when 5 students have seen the notes, whichever comes first. If exactly one student sees the notes every 15 seconds, what is the expectation and variance of the number of seconds that the experiment lasts?

Answer 1

let experiment runs till X student.

below is pmf of X:

P(X=1)=P(1st student sees and copy) =1/6

P(X=2)=P(!st does not see and 2nd see and copy)= (1-1/6)*(1/6)=5/36

P(X=3)=25/216

P(X=4)=125/1256

P(X=5)=P(none of first 4 sees and copy)=(5/6)^4 =625/1296

E(x)=ΣxP(x)=1*1/6+2*5/36+3*25/216+4*125/1256+5*625/1296=3.5887

E(x²) =Σx²P(x)=1²*1/6+2²*5/36+3²*25/216+4²*125/1256+5²*625/1296=15.3634

Var(x)=E(x²) -(E(X))² =2.4844

since time of experiment Y =15X

therefore

expectation=E(15X) =15*3.5887 =53.83 seconds

and variance =Var(15X) =15^2*Var(x) =225*2.4844 =558.99 seconds²