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 January 28th, 2015, 12:54 AM #1 Newbie   Joined: Jan 2015 From: Spain Posts: 1 Thanks: 0 probablity - n previolusly persons failed exam (tricky) We have an exam. Students are staying in queue. After every student probablity that professor finish exam is $\frac12$. For first student in queue there is $\frac12$ probablity that he pass. When first student fails exam then second student with probabality $\frac{1}{2+2}$ fail exam. And generally $k-th$ student fails exam with probablity $\frac{1}{j+2}$ where $j$ is number of students that failed exam. So potentially the best idea is to be last in the queue, but in this case we have the biggest probablity that proffessor finish exam before we try pass. Find positions in queue that our probablity to pass is $\ge 1/2$ So look at my attempt: We get $k$-th postiotn. Probablity that professor don't finish is $\frac{1}{2^{k-1}}$ And now let $N_0$ - 0 students failed exam $N_1$ - one student failed exam $...$ $N_{k-1}$ - $(k-1)$ students failed exam $A$ - $k$-th student didn't failed exam. So $$P(A)=\frac{1}{2^{k-1}}\sum_{i=0}^{k-1}P(A|N_i)P(N_i) = \frac{1}{2^{k-1}}\sum_{i=0}^{k-1}\frac{1}{2i+2}P(N_i)=\frac{1}{2^{k}}\sum_{i=1}^{ k}\frac{1}{i}P(N_{i-1})$$ Now my problem is to compute \$P(N_{i}). Could you help me, please ?

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