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Expert replies
by sudhir3127 » Thu Jul 24, 2008 3:32 am
If a paper based examination contains 150 questions and each question contains 4 options find the total number of ways in which a student can answer atleast 2 out of 150 questions.

OA after some discussion.
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Source: — Problem Solving |

by Saule » Thu Jul 24, 2008 5:45 am
Can you post all answer options?
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by Saule » Thu Jul 24, 2008 6:02 am
Is it 3 (1+2+3+...+148) ?
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by pepeprepa » Thu Jul 24, 2008 6:46 am
Total number of possibilities to answer the 150 questions: 4^150
Possibilities when 0 good answer: 3^150 // 3 possibilities for each question in order to be always wrong
Possibilities when 1 good answer: (3^149)*150 // 3 wrong possibilities for 149 questions and only 1 possibility to get the right one // The right solution can either be the first, the sedond or the 150th question, so we multiply by 150.

Possibilities when 2 good answers or more than 2:
4^150 - ( 3^150 + (3^149)*150)= 4^150 - ( 3^150*(1+50) )
=4^150 - 51 * 3^150


4^150 - 51 * 3^150 is the total number of ways in which a student can answer atleast 2 out of 150 questions
Last edited by pepeprepa on Thu Jul 24, 2008 6:54 am, edited 1 time in total.
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by parallel_chase » Thu Jul 24, 2008 6:54 am
Is the answer 3776?
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by sudhir3127 » Thu Jul 24, 2008 6:59 am
Here goes the Answer...

A student may answer a question or he may not .So the total chances that a student can take with a question.

Consider question no.1

one can take 5 chances with it namely answering either option A or B or C or D or can leave it out.

So with each and every question a student can take 5 chances.

So total no. of chance is 5^150.

From this we have to subtract Total no of ways of answering 0 questions as well as 1 question exactly which is 1+ 150x4 = 601

So the final answer is 5^ 150 - 601
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