Question

Students sometimes delay doing laundry until they finish their problem sets. Assume all random values described below are mutually independent.

(a)A busy student must complete 3 problem sets before doing laundry. Each problem set requires 1 day with probability 2/3 and 2 days with probability 1/3. Let B be the number of days a busy student delays laundry. What is Ex(B)? Example: If the first problem set requires 1 day and the second and third problem sets each require 2 days, then the student delays for B=5 days.

(b)A relaxed student rolls a fair,6-sided die in the morning. If he rolls a1, then he does his laundry immediately (with zero days of delay). Otherwise, he delays for one day and repeats the experiment the following morning. Let R be the number of days a relaxed student delays laundry. What is Ex(R)? Example:If the student rolls a 2 the first morning, a 5 the second morning, and a 1 the third morning, then he delays for R=2 days.

(c)Before doing laundry, an unlucky student must recover from illness for a number of days equal to the product of the numbers rolled on two fair,6-sided dice. Let U be the number of days an unlucky student delays laundry. What is Ex(U)? Example: If the rolls are 5 and 3, then the student delays for U=15 days.

(d)A student is busy with probability 1/2, relaxed with probability 1/3, and unlucky with probability1/6. Let D be the number of days the student delays laundry. What is Ex(D)?

Answer 1

(a)P( a problem set requires 1 day)=2/3

P( a problem set requires 2 days)=1/3

no. of problems a busy student must solve before laundry=3

B= no. of days a busy student delay laundry

then,

=1*2/3+2*1/3

=0.66+0.66 = 1.32 days

(b) R= no. of days a relaxed student delays laundry

P( a fair die will show 1)=1/6

P( a fair die will not show 1)=1-1/6= 5/6

so, P(a relaxed student delays laundry)=5/6

E(R)=2*1/5+3*1/5+4*1/5+5*1/5+6*1/5

=1/5(2+3+4+5+6)

=1/5(20)

=4 days