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Re: A school examination consists of five questions, each having two parts [#permalink]
5 questions each having 2 ways so We have 10 Answers. Number of ways are always 2^(n).
2^10=1024

But in 1 way all questions will be left blank. So reduce that >>1024-1= 1023
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Re: A school examination consists of five questions, each having two parts [#permalink]
RishS94 wrote:
5 questions each having 2 ways so We have 10 Answers. Number of ways are always 2^(n).
2^10=1024

But in 1 way all questions will be left blank. So reduce that >>1024-1= 1023

­didn't quite understand how we arrived at 2^10,  since there are 10 question the total number of ways should be 10!
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A school examination consists of five questions, each having two parts [#permalink]
Pranitashetty For each question, we have two possibilities: to attempt or not. By the fundamental counting principle, we have 2 possibilities for each question. That is 2 times 2 ten times, representing the possible ways for the student to give it a try. He might, for example, answer both parts of question one or just the first part of question five. Or he could even choose not to answer any of them. So there are 2^10 possible ways for him to approach the test, 2 for each of the 10 questions, multiplied because of the fundamental counting principle.

2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^10

The question asks how many ways he can attempt at least one, which is all the possibilities minus the 1 in which he attempts none.

(2^10)−(1) = 1024 - 1 = 1023
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Re: A school examination consists of five questions, each having two parts [#permalink]
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­HI Experts,

Doesnt "Atleast one" mean : One or more?? How do we figure that atleast one meaning the student will not attempt any of the questions from 10? Kindly clarify.
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A school examination consists of five questions, each having two parts [#permalink]
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arshavishah11 wrote:
­HI Experts,

Doesnt "Atleast one" mean : One or more?? How do we figure that atleast one meaning the student will not attempt any of the questions from 10? Kindly clarify.

­Yes, it does! The probability we're looking for is the probability of attempting one or more questions. So, we need to consider all possibilities except the one where no part of any question is attempted. To find this, we count all possibilities and subtract the one that does not satisfy the condition (where no part is attempted). We already know that there is olny one way to not attempt any question.

For each question, there are two possibilities: to attempt it or not. This gives us:

2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^10 possibilities,

which equals 1024 possibilities. Subtracting the one scenario where no question is attempted:

1024 - 1 = 1023

So, there are 1023 possibilities where one or more part of a question was attempted.­
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A school examination consists of five questions, each having two parts [#permalink]
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