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anonymous
 one year ago
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
anonymous
 one year ago
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

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hartnn
 one year ago
Best ResponseYou've already chosen the best response.2so he clears 2 exams or all 3 in 2 exam's case, there are 3 possibilities DI, VA VA,QA DI,QA

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2In 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. But if when I add, the answer comes out to be 0.564

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The answer is 0.0402

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2let me try another way 1 [ (1 0.2)*(10.3) + (1 0.2)*(10.9)+(1 0.9)*(10.3)] that comes out to be 0.29 :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is there any other method?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Probability for clearing all 3 : \(0.2*0.3*0.9 \) Probability for clearing exactly 2 : \( (10.2)*0.3*0.9 + 0.2*(10.3)*0.9 + 0.2*0.3*(10.9)\) Adding them gives http://www.wolframalpha.com/input/?i=0.2*0.3*0.9+%2B++%2810.2%29*0.3*0.9+%2B++0.2*%2810.3%29*0.9+%2B+0.2*0.3*%2810.9%29

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2aah! didn't consider, exactly 2 case! passing in exactly 2 is (1a)*b*c and not just b*c

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you @ganeshie8
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