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 June 21st, 2019, 10:04 AM #1 Newbie   Joined: Jun 2019 From: Melbourne Posts: 4 Thanks: 0 Probability "A lecturer teaches Digital Communication to 30 of her students. However she doesn't know if her student grasp her lecture. The students have to raise their hands if they dont grasp the lecture. She gives dice to a student and asks him to roll it. She does it with every student in the class. Following are the conditions 1)If 1 shows up, then the student honestly raises his/her hand if he/she did not understand the concept 2)If 2,3,5 show up then the student raises his/her hand irrespective of understanding the concept 3)If 4 or 6 shows up then student does not raise his/her hand. Let X~Bernoulli(q) where student raises his/her hand and Y~Bernoulli(p) where a student understands the concept. We can write q=Ap+B. Determine A and B."
 June 21st, 2019, 12:32 PM #2 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 911 Thanks: 64 Math Focus: सामान्य गणित $\displaystyle q = \frac {-1}{6}p +\frac {2}{3}$ I got this
 June 22nd, 2019, 05:19 PM #3 Newbie   Joined: Jun 2019 From: Melbourne Posts: 4 Thanks: 0 Can you please elaborate your working please
 June 22nd, 2019, 05:45 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,649 Thanks: 1476 $q=P[X] = P[1]p + P[2,3,5] = \dfrac{p}{6} + \dfrac 1 2$ $A=\dfrac 1 6,~B=\dfrac 1 2$ Thanks from kallurayaankit
 June 23rd, 2019, 12:21 AM #5 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 911 Thanks: 64 Math Focus: सामान्य गणित Let $\displaystyle p$ be the probability of understanding the concept. $\displaystyle (1-p)$ = probablity of not undersranding the concept. Since, 2, 3, 5 will always raise their hands. 4, 6 won't raise their hands. 1 will raise only if they dont understand. Probability of hands being raised = $\displaystyle \frac {1}{2} + 0 + \frac {1}{6}(1-p) = \frac {-1}{6}p + \frac {2}{3 }$

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