Hi,
Each question can be either T or F. So, total number of possible answers is 2^8
For seven questions to be right, 1 option should be wrong. So, any of the wrong questions can be made incorrect by choosing the opposite option (i.e T instead of F and vice-versa) in 1 way
Question that can be made incorrect can be chosen in 8C1 ways.
So, probability is 8C1.1/2^8 = 1/32

Hence, A

Frankenstein wrote:Hi,
Each question can be either T or F. So, total number of possible answers is 2^8
For seven questions to be right, 1 option should be wrong. So, any of the wrong questions can be made incorrect by choosing the opposite option (i.e T instead of F and vice-versa) in 1 way
Question that can be made incorrect can be chosen in 8C1 ways.
So, probability is 8C1.1/2^8 = 1/32

Hence, A

Thanks Frankesnstein for ur help. I don't get the highlighted part. I get that we have 2^8 answer choices to choose from. We need to know the probability of getting exactly 7 right answers out of, can you explain this part? Thanks

Okay.. Out of those 8 questions we have to make 1 question incorrect right?
That incorrect question can be either 1st,2nd, 3rd ...8th right?
So, we can make 7 questions right in 8 ways right?
For example:
If the answer key for the question paper is T,T,T,T,T,T,T,T
We can have 7 questions correct in 8 ways:
1)F,T,T,T,T,T,T,T
2)T,F,T,T,T,T,T,T
3)T,T,F,T,T,T,T,T
4)T,T,T,F,T,T,T,T
5)T,T,T,T,F,T,T,T
6)T,T,T,T,T,F,T,T
7)T,T,T,T,T,T,F,T
8)T,T,T,T,T,T,T,F

That is essentially picking 1 question from the 8 questions to make it incorrect - 8C1 ways

so basically the probability of getting 7 right is the probability of getting 1 wrong and this can happen in 8 ways, right? so out of the 2^8 possible answers, there're 8 ways an answer can be wrong or 7 can be right.

8/2^8 = 1/2^5

Thanks for explaining.