Question

A dime is tossed repeatedly until a head appears. Let N be the trial number on which this first head occurs. Then a nickel is tossed N times. Let X count the number of times that the nickel comes up tails. Determine Pr{X = 0) and E[X].

Answer 1

Answer:

P(X=0) = 1 , E(X)=0

Step-by-step explanation:

If the dime is balanced , then defining the variable Y= number tails until the first head occurs , then Y follows a negative binomial distribution , with parameter k=N-1 and r=1 (N-1 successes until the r-th failure).

Then

P(Y)= C(k+r-1,k)*(1-p)^r*p^k

P(Y)=C(N-1,N-1) * (1-p)^1 * p^(N-1)

P(Y)= (1-p)*p^(N-1)

where

C ( ) = combinations

p= probability of obtaining heads for every toss = 0.5

P(Y)= probability that there are N-1 tails and 1 head at last → we set P(Y)= 0.999

then

0.99= 0.5*0.5^(N-1)

0.999= 0.5^N

thus

N= ln(0.999)/ln(0.5) = 1.443*10⁻³

while X= number of times that the nickel comes up tails with N trials , follows a binomial distribution

P(X=x)= C(N,x)*(1-p)^(N-x)*p^x

for x=0

P(X=0)= C(N,0)*(1-p)^(N-0)*p^0 = 1*(1-p)^N *1 = (1-p)^N

since p=1-p=0.5

P(X=0)=0.5^N = 0.999

and the expected value of X can be calculated through

E(X)=N*p= 1.443*10⁻³*0.5 = 7.217*10⁻⁴

if we take a better approximation , taking P(Y) closer to 1 then P(X=0)→ 1 and E(X)→0