Question

There are 40 students in a Probability Course. Before epidemics their professor did his exams in campus and to avoid cheating cases he used two variants of the quiz by printing 20 papers of A variant and 20 papers of B variant. When the quiz ended, the students randomly put them into one pile. When the professor returned to his office he started to sort them into two piles of A variant and B variant. He took each paper from top of the original pile and put into one of the two piles A and B. Each time he switched from pile A to pile B or vice versa he took a deep breath. In average how many times did he take deep breath during this sorting process? For example, in the order ABBAABAB he takes 5 deep breaths: A*BB*AA*B*A*B (* indicates deep breath).

Answer 1

Answer:

Let X be the random variable that denotes the number of deep breaths taken by the professor. The range of X is from 1 to 40. He can either take a minimum of one deep breath by sorting the 20 papers from one variant first and then the remaining papers from the other variant like AAA...AA*BBB...BB or he can take a maximum of 39 deep breaths by selecting one paper from a particular variant at every alternate pick like A*B*A*B*A*B*A*B... and so on. We are required to compute the expected value of X, that is, E(X).

According to the given problem, X can take exactly two values 0 or 1 where 1 represents the professor taking a deep breath, that is, picking up a paper from the other variant while 0 represents the professor not taking a deep breath, that is he picked up a paper from the same variant. X taking value 1 is considered as a succes while taking the value 0 is considered as a failure. The probability of success is = probability of choosing one of the two variants' papers = 1/2 = p and the number of trials of the experiment is one less than the total number of papers from the two variants = 40-1 = 39 = n.

Hence, the random variable X follows a binomial distribution with parameters : number of trials, n =40 and probability of success, p = 1/2, that is, X ~ Binomial(39,0.5)

On average, the number of times he took deep breaths during this sorting process is nothing but the expected value of X = E[X]. Now, for a binomial distribution with paramters n and p we know its expectation = n*p

Therefore, E[X] = n*p = 39*0.5 = 19.5

Hence, on an average the professor took around 20 deep breaths during this sorting process.

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