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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!

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 ?

Geometric Distribution

Expectation of geometric distribution

Basic counting

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 !

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 !!

What to do to shape your Career in Mathematics after 12th?

From the video below, let's learn from Dr. Ashani Dasgupta (a Ph.D. in Mathematics from the University of Milwaukee-Wisconsin and Founder-Faculty of Cheenta) how you can shape your career in Mathematics and pursue it after 12th in India and Abroad. These are some of the key questions that we are discussing here:

- What are some of the best colleges for Mathematics that you can aim to apply for after high school?
- How can you strategically opt for less known colleges and prepare yourself for the best universities in India or Abroad for your Masters or Ph.D. Programs?
- What are the best universities for MS, MMath, and Ph.D. Programs in India?
- What topics in Mathematics are really needed to crack some great Masters or Ph.D. level entrances?
- How can you pursue a Ph.D. in Mathematics outside India?
- What are the 5 ways Cheenta can help you to pursue Higher Mathematics in India and abroad?

Cheenta has taken an initiative of helping College and High School Passout Students with its "Open Seminars" and "Open for all Math Camps". These events are extremely useful for students who are really passionate for Mathematic and want to pursue their career in it.

Cheenta is a knowledge partner of Aditya Birla Education Academy

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