OpenStudy (anonymous):

Shamil is taking an exam which has 3 sections,DI,VA and QA.He identifies that his probabilities of clearing cut offs in these sections are 0.2, 0.3, and 0.9 respectively. If these cut offs are independent of each other , find the probability that Shamil clears at least two cut offs. 1) 0.622 2) 0.484 3) 0.042 4) 0.402

3 years ago
hartnn (hartnn):

so he clears 2 exams or all 3 in 2 exam's case, there are 3 possibilities DI, VA VA,QA DI,QA

3 years ago
hartnn (hartnn):

In probability, when you see OR, you ADD< when you see AND, you multiply (DI and VA) or (VA and QA) or (DI and QA) or (DI and VA and QA) making sense?

3 years ago
OpenStudy (anonymous):

Yes. But if when I add, the answer comes out to be 0.564

3 years ago
hartnn (hartnn):

yea, i too get that

3 years ago
OpenStudy (anonymous):

The answer is 0.0402

3 years ago
hartnn (hartnn):

let me try another way 1- [ (1- 0.2)*(1-0.3) + (1- 0.2)*(1-0.9)+(1- 0.9)*(1-0.3)] that comes out to be 0.29 :P

3 years ago
OpenStudy (anonymous):

Is there any other method?

3 years ago
ganeshie8 (ganeshie8):

Probability for clearing all 3 : \(0.2*0.3*0.9 \) Probability for clearing exactly 2 : \( (1-0.2)*0.3*0.9 + 0.2*(1-0.3)*0.9 + 0.2*0.3*(1-0.9)\) Adding them gives http://www.wolframalpha.com/input/?i=0.2*0.3*0.9+%2B++%281-0.2%29*0.3*0.9+%2B++0.2*%281-0.3%29*0.9+%2B+0.2*0.3*%281-0.9%29

3 years ago
hartnn (hartnn):

aah! didn't consider, exactly 2 case! passing in exactly 2 is (1-a)*b*c and not just b*c

3 years ago
OpenStudy (anonymous):

Thank you @ganeshie8

3 years ago