Question

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

Answer 1

so he clears 2 exams or all 3

in 2 exam's case, there are 3 possibilities
DI, VA
VA,QA
DI,QA

Answer 2

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?

Answer 3

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

Answer 4

yea, i too get that

Answer 5

The answer is 0.0402

Answer 6

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

Answer 7

Is there any other method?

Answer 8

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

Answer 9

aah!

didn't consider, exactly 2 case!

passing in exactly 2 is
(1-a)*b*c

and not just b*c

Answer 10

Thank you @ganeshie8

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