probability problem...kinda tricky

BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

This topic has expert replies

Post new topic Post Reply

Previous Topic Next Topic

knight247: Legendary Member; Posts: 504; Joined: Tue Apr 19, 2011 1:40 pm; Thanked: 114 times; Followed by:11 members

probability problem...kinda tricky

by knight247 » Thu Jun 23, 2011 5:54 am

In a quiz, a contestant is asked 10 questions, to which he should answer true or false. He should answer at least 8 questions correctly to move to the next level of the quiz. If he answers all the questions, what is the probability that he does not proceed to the next level?
(A)7/128
(B)17/28
(C)121/128
(D)11/28
(E)10/128

This problem has been bugging me all day. A detailed explanation would be highly appreciated. OA is C

Quote

Source: Beat The GMAT — Problem Solving | Thu Jun 23, 2011 5:54 am

Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 am; Location: India; Thanked: 375 times; Followed by:53 members

by Frankenstein » Thu Jun 23, 2011 6:00 am

Hi,
Each question can be either T or F.
So, total number of different sets of answers is 2^10
Number of ways of getting 8 correct is choosing 8 correct options from 10, i.e. 10C8 = 45
Number of ways of getting 9 correct is choosing 9 correct options from 10, i.e. 10C9 = 10
Number of ways of getting all correct is = 1
So, probability of passing is (1+10+45)/2^10 = 56/1024 = 7/128
So, probability that he does not proceed is 1- (7/128) = 121/128

Hence, C

Cheers!

Things are not what they appear to be... nor are they otherwise

Quote

finites: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Thu Jun 02, 2011 7:21 am

by finites » Thu Jun 23, 2011 10:17 am

Hi Frank,

I understood that there 2^10 possible answers..

Now what i have NOT understood is how 10C8 is the way of getting 8 correct answers.

For each Question the answer might be true or False, both have equal probability to become correct.

In my opinion 10C8 is way of answering 8 questions out of 10.

I have a question of how this is the way of getting correct answers for 8 questions.. Please help ..

Frankenstein wrote:Hi,
Each question can be either T or F.
So, total number of different sets of answers is 2^10
Number of ways of getting 8 correct is choosing 8 correct options from 10, i.e. 10C8 = 45
Number of ways of getting 9 correct is choosing 9 correct options from 10, i.e. 10C9 = 10
Number of ways of getting all correct is = 1
So, probability of passing is (1+10+45)/2^10 = 56/1024 = 7/128
So, probability that he does not proceed is 1- (7/128) = 121/128

Hence, C

Quote

Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 am; Location: India; Thanked: 375 times; Followed by:53 members

by Frankenstein » Thu Jun 23, 2011 10:27 am

finites wrote:Hi Frank,

I understood that there 2^10 possible answers..

Now what i have NOT understood is how 10C8 is the way of getting 8 correct answers.
For each Question the answer might be true or False, both have equal probability to become correct.
In my opinion 10C8 is way of answering 8 questions out of 10.
I have a question of how this is the way of getting correct answers for 8 questions.. Please help ..

Hi,
Let me elaborate this:
A question can be made correct in only 1 way right?
Similarly a question can be made incorrect in only 1 way right?
Now, let us pick 8 questions. These 8 questions will be correct in only 1 way right(all those 8 questions should match the key)?
Now, these 8 questions can be picked from 10 in 10C8 ways right?
So, the number of ways of making 8 questions right is (Number of ways of picking 8 questions)*(number of ways of making those 8 questions right) = 10C8*1 = 10C8.
If you still feel I haven't explained well, let me know at which specific point you didn't understand so that I can explain better.

Cheers!

Things are not what they appear to be... nor are they otherwise