goformit100
  • goformit100
In an examination there are three multiple choice questions and each question has 4 choices. Number of ways in which a student can fail to get all answer correct is (A) 11 (B) 12 (C) 27 (D) 63
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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goformit100
  • goformit100
@dan815
dan815
  • dan815
total prob - total prob of all correct
dan815
  • dan815
oh ways total ways - total ways to get all correct

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More answers

goformit100
  • goformit100
How can I get the explanation ?
dan815
  • dan815
no thats not right
dan815
  • dan815
3 multiple choice so
dan815
  • dan815
|dw:1371199600693:dw|
dan815
  • dan815
63
dan815
  • dan815
|dw:1371199680103:dw|
dan815
  • dan815
so 64-1
goformit100
  • goformit100
ok
anonymous
  • anonymous
For the total possible ways to answer, we find there are \(4\) possible answers and \(3\) questions so \(4\times4\times4=4^3=64\). For the number of ways to *not* answer all correctly, consider the number of ways to answer all correctly, first. It should be intuitive there is only \(1\) way to answer everything correctly, so \(64-1=63\) is then the rest of our possible ways of answering which result in *not* answering all correctly.
DLS
  • DLS
Q1) a b c d Q2) a b c d Q3) a b c d Let us assume correct answers are abc repsectively..then No.of ways for incorrect attempts: well.. then simply, |dw:1371199806943:dw| does this make sense?
goformit100
  • goformit100
Thanks
DLS
  • DLS
yope,63 hi-fi @oldrin.bataku :"D

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