Get inspired by the success stories of our students in IIT JAM MS, ISI  MStat, CMI MSc DS.  Learn More 

ISI MStat PSB 2013 Problem 9 | Envelope Collector's Expenditure

This is a very simple and beautiful sample problem from ISI MStat PSB 2013 Problem 9. It is mainly based on geometric distribution and its expectation . Try it!

Problem- ISI MStat PSB 2013 Problem 9


Envelopes are on sale for Rs. 30 each. Each envelope contains exactly one coupon, which can be one of the one of four types with equal probability. Suppose you keep on buying envelopes and stop when you collect all the four type of coupons. What will be your expenditure ?

Prerequisites


Geometric Distribution

Expectation of geometric distribution

Basic counting

Solution :

This problem seems quite simple and it is simple, often one may argue that we can take a single random variable, which denotes the number of trials till the fourth success (or is it third !!), and calculate its expectation. But I differ here becaue I find its lot easier to work with sum of geometric random variables than a negative binomial. ( negative binomial is actually sum of finite geometrics !!)

So, here what we will do, is define 4 random variables, as X_i : # trials to get a type of coupon that is different from the all the (i-1) types of coupons drawn earlier. i=1,2,3,4.

Now since each type of coupon has an equal probability to come, that is probability of success is \frac{1}{4}, here a common mistakes people commit is assuming all X_1,X_2,X_3,X_4 are i.i.d Geometric(\frac{1}{4}), and this turns out to be a disaster !! So, be aware and observe keenly, that at the first draw, any of the four types will come, with probability 1, and there after we just need the rest of the 3 types to appear at least once. So, here X_1=1 always and X_2 \sim Geo(\frac{3}{4})( becuase since in the first trial, surely any of the four types will come, our success would be getting any of the 3 types of envelopes from the all types making the success probability \frac{3}{4}) similarly , X_3 \sim Geo(\frac{1}{2}) and X_4 \sim Geo(\frac{1}{4}).

Now for the expectated expenditure at the given rate of Rs. 30 per envelope, expected expenditure is Rs.30 E(X_1+X_2+X_3+X_4)

Now, we know that E(X_2)=\frac{4}{3}, E(X_3)=2 and E(X_4)=4 (why??)

So, E(X_1+X_2+X_3+X_4)=1+\frac{4}{3}+2+4=\frac{25}{3}.

So, our required expectation is Rs. 250. Hence we are done !


Food For Thought

Suppose among the envelopes you collected each envelope has an unique 5 digit number, you picked an envelope randomly and what is the chance of the number that is going to show up is has its digits in a non-decreasing order ?

How does the chances may vary if you find a k digit number on the envelope ? think over it !!


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


This is a very simple and beautiful sample problem from ISI MStat PSB 2013 Problem 9. It is mainly based on geometric distribution and its expectation . Try it!

Problem- ISI MStat PSB 2013 Problem 9


Envelopes are on sale for Rs. 30 each. Each envelope contains exactly one coupon, which can be one of the one of four types with equal probability. Suppose you keep on buying envelopes and stop when you collect all the four type of coupons. What will be your expenditure ?

Prerequisites


Geometric Distribution

Expectation of geometric distribution

Basic counting

Solution :

This problem seems quite simple and it is simple, often one may argue that we can take a single random variable, which denotes the number of trials till the fourth success (or is it third !!), and calculate its expectation. But I differ here becaue I find its lot easier to work with sum of geometric random variables than a negative binomial. ( negative binomial is actually sum of finite geometrics !!)

So, here what we will do, is define 4 random variables, as X_i : # trials to get a type of coupon that is different from the all the (i-1) types of coupons drawn earlier. i=1,2,3,4.

Now since each type of coupon has an equal probability to come, that is probability of success is \frac{1}{4}, here a common mistakes people commit is assuming all X_1,X_2,X_3,X_4 are i.i.d Geometric(\frac{1}{4}), and this turns out to be a disaster !! So, be aware and observe keenly, that at the first draw, any of the four types will come, with probability 1, and there after we just need the rest of the 3 types to appear at least once. So, here X_1=1 always and X_2 \sim Geo(\frac{3}{4})( becuase since in the first trial, surely any of the four types will come, our success would be getting any of the 3 types of envelopes from the all types making the success probability \frac{3}{4}) similarly , X_3 \sim Geo(\frac{1}{2}) and X_4 \sim Geo(\frac{1}{4}).

Now for the expectated expenditure at the given rate of Rs. 30 per envelope, expected expenditure is Rs.30 E(X_1+X_2+X_3+X_4)

Now, we know that E(X_2)=\frac{4}{3}, E(X_3)=2 and E(X_4)=4 (why??)

So, E(X_1+X_2+X_3+X_4)=1+\frac{4}{3}+2+4=\frac{25}{3}.

So, our required expectation is Rs. 250. Hence we are done !


Food For Thought

Suppose among the envelopes you collected each envelope has an unique 5 digit number, you picked an envelope randomly and what is the chance of the number that is going to show up is has its digits in a non-decreasing order ?

How does the chances may vary if you find a k digit number on the envelope ? think over it !!


Similar Problems and Solutions



ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
Menu
Trial
Whatsapp
rockethighlight