lecture-4 - Lecture 4 Random variable and expected value 8...

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Unformatted text preview: Lecture 4 Random variable and expected value 8 Jan, 2008 In the last two lectures, we discussed elementary probability theory. We also mentioned a few proposi- tions and proved a few Theorems. However, till now we have not seen any better way to calculate probability than an exhaustive way of calculating the probability in which we calculate probability of an event by sum- ming up the probability of its elementary events. For example, if we toss a fair coin n times, what is the probability of seeing i heads. Here the sample space consists of 2 n elementary events and due to fairness of coin, each elementary event is equally likely. In other words probability of each elementary event is 1 / 2 n . Now to calculate the probability of the desired event ” i heads occur in n tosses”, it just suffices to count the elementary events having i heads and n- i tails. This count is clearly ( n i ) . Hence the probability of seeing i heads in n throws of a fair coin is ( n i ) / 2 n . We are very often interested in some quantitative measure other than just computing the probability of an event and in such situation the above naive way of calculating probability is sometimes very inefficient. For example, consider the following problems. • Experiment : We throw r balls uniformly randomly and independently into n bins. Question : What is the average number of empty bins ? To solve this problem using the naive way would require calculating the probability that m bins are empty, multiplying it by m , and summing these product terms for all m < n . Calculating the probability that m bins are empty was left as an exercise, and you might have realized that there is a messy formula involving double summation operators. As a surprise, there is a closed form expression for the answer : n (1- 1 /n ) r . This motivates use to explore some better and calculation-free way to solve this problem. • Experiment : There is a bin containing r red balls and b black balls. We take out these balls one at a time and place then in the order they are taken out. At each step, each of the ball in the bin is equally likely to be taken out. Question : What is the average number of red balls which precede all black balls ? Let us consider a small instance of the above problem. Suppose there are 3 red balls and 2 black balls in a bin. We can solve this problem quite easily as follows. For each elementary event, find out the ”the number of red balls preceding all black balls”, multiply it with the probability of that elementary event and sum it up for all events in the sample space. We may be bit clever in doing so by partitioning the sample space in to groups such that ”the number of red balls preceding all black balls” is same...
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This note was uploaded on 04/28/2009 for the course MATH 332 taught by Professor Sicha during the Spring '05 term at Accreditation Commission for Acupuncture and Oriental Medicine.

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lecture-4 - Lecture 4 Random variable and expected value 8...

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