Question

5. An Aggie ECE student attends a programming competition that consists of an endless se- quence of tasks the student must write a code for. Given any task, independently of other tasks, the student writes a code with no error with probability,or writes a code with minor errors with probability 3or fails the task with probability The student gets 1 point, 0.5 point, or 0 point for every code with no error, minor errors, or a failed code, respectively Let N and M be respectively the number of codes that the student writes with no error and minor errors until the first failure occurs. (a) 1 pt Find the conditional PMF of M given N 1. Let S be the total score (in points) that the student gets until the first failure. (b) 1 pt Find the conditio ( nal PMF of M given S 1. (c) 0.75 pt Compute the conditional mean of M given N 1 (d) 0.75 pt Compute the conditional variance of M given N 1. (Hint: For a geometric random variable X with parameter p, we have EX-6- 2)

Answer 1

Here the events M and N as specified in the problem and fails the task are independent to one another.

(a) The conditional PMF of M given N = 1 is equal to PMF of M becuse basically M and N are independent.

Therefore PMF of M is P(M = x minor errors before the first failure) = (1/8)^x(1/8) = (1/8)^x+1 for x = 0, 1, 2,....

(b) The conditional PMF of M given S = 1 is equal to

P ( M = x minor errors before the first failure $\cap$ S = 1) / P(S = 1)

{Here, S gets total 1 in any one of the two cases as (one N and failure ) or (Two Ms and failure) with probability as P(S = 1) = (3/4)(1/8) + (1/8)²(1/8)}

=  P ( M = x minor errors before the first failure $\cap$ S = 1) / P(S = 1)

=P ( M = x minor errors before the first failure $\cap$ (one N and failure ) or (Two Ms and failure)) / {(3/4)(1/8) + (1/8)²(1/8)}

= P ( Two Ms = x minor errors before the first failure) / {(3/4)(1/8) + (1/8)²(1/8)}

= (1/8)^2x(1/8) / {(3/4)(1/8) + (1/8)²(1/8)}

= (1/8)^2x+1 / {(3/4)(1/8) + (1/8)²(1/8)} for x = 0, 1, 2, .....

(c) Since M and N are independent and also M follows geometric distribution.

Then conditional mean of M given N = 1 is equal to the un conditional mean of M using geometric distribution

is given by P(M) / P(failure) = (1/8)/(1/8) = 1 .

(d) Since M and N are independent and also M follows geometric distribution.

Then conditional variance of M given N = 1 is equal to the un conditional variance of M using geometric distribution  is given by P(M) / [P(failure)]² = (1/8)/(1/8)² = 8.