P&C

[sudhir3127]

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.

[Saule]

Can you post all answer options?

[Saule]

Is it 3 (1+2+3+...+148) ?

[pepeprepa]

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

[parallel_chase]

Is the answer 3776?

[sudhir3127]

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