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].

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