A school examination consists of five questions, each having two parts

Question

A school examination consists of five questions, each having two parts. In how many ways can a student attempt at least one part of any of the questions?

(A) 624
(B) 625
(C) 1023
(D) 1024
(E) 4095

whatsarc · Answer

Each of the 5 questions has 2 parts, so 10 parts in total.

Total number of ways to attempt the parts is 2^10.

There is only 1 way to not attempt any parts.

Total number of ways to attempt  minus  the number of ways to not attempt

= 2^10 - 1
= 1024-1
= 1023

Dhruv805 · Answer

can you give an in detail explanation pls

RishS94 · Answer

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

Pranitashetty · Answer

didn't quite understand how we arrived at 2^10,  since there are 10 question the total number of ways should be 10!

FelipeDiaz · Answer

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

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

arshavishah11 · Answer

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.

FelipeDiaz · Answer

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.

A school examination consists of five questions, each having two parts

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