A bizarre reality!

Probability Level 4

Out of a bunch of $n$ hand-outs a professor wants to give to his students as background to his lecture, $k$ have typographic errors.Each student receives one hand-out as he/she enters the classroom.Let three students Ajay,Vijay and Sujay be our participants for a probability experiment.The trio enter the classroom one after the other.Let $p_{ajay}$ denote the probability of Ajay receiving a faulty hand-out.Similarly,let $p_{vijay}$ and $p_{sujay}$ respectively denote the probabilties of them getting a faulty hand-out.Then,

p_{ajay}=p_{vijay}=p_{sujay}

for all permissible values of

n

and

k

p_{ajay}=p_{vijay}=p_{sujay}

only when

k \le n-k

p_{ajay}=p_{vijay}=p_{sujay}=\frac{1}{2}

for all permissible

(n,k)

p_{ajay}=p_{vijay}=p_{sujay}

is not possible for any

(n,k)

p_{ajay}=p_{vijay}=p_{sujay}

only when

k \le n-4k

p_{ajay}=p_{vijay}=p_{sujay}

only when

5k=n

1 solution

Aryan Goyat
Apr 19, 2017

P(1stperson)=k/n

P(2ndperson)=k/n (k-1)/n+(1-k/n) k/n=k/n

so its independent of order by which they enter and a wow application ques !!!!!!!!!!

I think, there is error in your expression for P(second person).Note that P(2nd person) should be:

${(\frac{k}{n})}{(\frac{k-1}{n-1})} + (1-\frac{k}{n})(\frac{k}{n-1})$

Indraneel Mukhopadhyaya - 4 years, 1 month ago

A bizarre reality!

1 solution

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