An exam consists of 8 true/ false questions. Brian forgets to study, so he must guess blindly on each question. If any score above 70% is a passing grade, what is the probability that Brian passes?
A: 1/16 B: 37/256 C: ½ D:219/256 E: 15/16
OA D
An exam consists of 8 true/ false questions
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To pass, Brian must correctly answer at least 70% of the 8 questions:gmatdriller wrote:An exam consists of 8 true/ false questions. Brian forgets to study, so he must guess blindly on each question. If any score above 70% is a passing grade, what is the probability that Brian passes?
A: 1/16 B: 37/256 C: ½ D:219/256 E: 15/16
OA D
(7/10)(8) = 5.6.
Since the number of correctly answered questions must be an INTEGER VALUE, Brian will pass if he answers correctly at least 6 questions.
P = (good options)/(all possible options).
All possible options:
For each of the 8 questions, there are 2 options: correct or incorrect.
To combine our 2 options for each question, we multiply:
2*2*2*2*2*2*2*2 = 256.
A time-saving tip:
nCr = nC(n-r).
Good options:
From 8 questions, the number of ways to choose 6 to be answered correctly = 8C6 = 8C(8-6) = 8C2 = (8*7)/(2*1) = 28.
From 8 questions, the number of ways to choose 7 to be answered correctly = 8C7 = 8C(8-7) = 8C1 = 8.
From 8 questions, the number of ways to answer all 8 correctly = 1.
Total good options = 28+8+1 = 37.
Resulting probability:
good/all = 37/256.
The correct answer is B.
If the OA is D, then the OA is incorrect.
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First there are 2^8 different ways to answer the quiz, or 256.
The ways that the score will be above 75% is if he answers 6, 7, or 8 correctly.
8 is the easy one because there is just 1 solution, all correct. ---- 1
For seven, there are 8 different questions that he could have answered wrong ---- 8
Finally, for 6, you can treat it as arrangements of the letters: TTTTTTFF. This is simply 8!/6!2! = 28.
28 + 8 + 1 = 37
The answer is indeed 37/256
The ways that the score will be above 75% is if he answers 6, 7, or 8 correctly.
8 is the easy one because there is just 1 solution, all correct. ---- 1
For seven, there are 8 different questions that he could have answered wrong ---- 8
Finally, for 6, you can treat it as arrangements of the letters: TTTTTTFF. This is simply 8!/6!2! = 28.
28 + 8 + 1 = 37
The answer is indeed 37/256
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Let's say that you have NO IDEA how to solve this question.gmatdriller wrote:An exam consists of 8 true/ false questions. Brian forgets to study, so he must guess blindly on each question. If any score above 70% is a passing grade, what is the probability that Brian passes?
A: 1/16 B: 37/256 C: ½ D:219/256 E: 15/16
OA D
The great thing about probability questions is that, even if you don't know how to answer them, you can often eliminate some answers by using your gut instincts alone.
As Mitch showed, Brian needs to correctly guess at least 6 of the 8 questions.
How likely does that FEEL?
Well, if should feel less than 50% likely, which means we can eliminate C, D and E.
Great, in 10 seconds, we're down to a 50-50 guess between A and B.
Guess A or B and move on knowing that you just added a lot of time to your "time bank" that you can devote to other questions.
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probability is so tricky.
I m able to find than Brian needs to give at least 6 good answers out of 8. & total number of ways = 2^8 = 256
so it can 6,2 OR 7,1 OR 8
but after that totally confused
but now I got another trick
PPPPPPFF OR PPPPPPPF OR PPPPPPPP
8!/6!2! + 8!/7!1! + 8!/8!
28 + 8 + 1 = 37
so 37/256
I m able to find than Brian needs to give at least 6 good answers out of 8. & total number of ways = 2^8 = 256
so it can 6,2 OR 7,1 OR 8
but after that totally confused
but now I got another trick
PPPPPPFF OR PPPPPPPF OR PPPPPPPP
8!/6!2! + 8!/7!1! + 8!/8!
28 + 8 + 1 = 37
so 37/256
WHat is wrong with this approach?
P(6 at least)= 1-P(not at least 6)
= 1-P(1 and 2 and 3 and 4 and 5 correct answers)
we have 5 correct answers with P= 1/2*1/2*1/2*1/2*1/2= 1/2^5 = 1/32
1-1/32 = 31/32
Why is my approach wrong?
P(6 at least)= 1-P(not at least 6)
= 1-P(1 and 2 and 3 and 4 and 5 correct answers)
we have 5 correct answers with P= 1/2*1/2*1/2*1/2*1/2= 1/2^5 = 1/32
1-1/32 = 31/32
Why is my approach wrong?
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There are a few issues with that solution.Zoser wrote:WHat is wrong with this approach?
P(6 at least)= 1-P(not at least 6)
= 1-P(1 and 2 and 3 and 4 and 5 correct answers)
we have 5 correct answers with P= 1/2*1/2*1/2*1/2*1/2= 1/2^5 = 1/32
1-1/32 = 31/32
Why is my approach wrong?
First, you have considered only one outcome in which Brian fails: he gets questions 1, 2, 3, 4, and 5 correct, and the rest incorrect.
What about getting questions 1, 2, 3, 4, and 6 correct, and the rest incorrect?
What about getting questions 3 and 8 correct, and the rest incorrect?
Etc.
Also, your calculation for getting questions 1, 2, 3, 4, and 5 correct (and the rest incorrect) is missing some parts.
P(1, 2, 3, 4, and 5 correct and 6, 7, and 8 incorrect) = P(#1 correct AND #2 correct AND #3 correct AND #4 correct AND#5 correct AND #6 INcorrect AND #7 INcorrect AND #8 INcorrect)
= P(#1 correct) x P(#2 correct) x P(#3 correct) x P(#4 correct) xP(#5 correct) x P(#6 INcorrect) x P(#7 INcorrect) x P(#8 INcorrect)
= 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2
= 1/256
Cheers,
Brent
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Since 70% of 8 is 5.6, Brian must answer at least 6 questions correctly in order to pass the exam.gmatdriller wrote: ↑Fri Jul 17, 2015 11:25 pmAn exam consists of 8 true/ false questions. Brian forgets to study, so he must guess blindly on each question. If any score above 70% is a passing grade, what is the probability that Brian passes?
A: 1/16 B: 37/256 C: ½ D:219/256 E: 15/16
OA D
The probability that he answers exactly 6 questions correctly is:
8C6 x (1/2)^6 x (1/2)^2 = 28 x (1/2)^8 = 28/256
The probability that he answers exactly 7 questions correctly is:
8C7 x (1/2)^7 x (1/2)^1 = 8 x (1/2)^8 = 8/256
The probability that he answers all 8 questions correctly is:
8C8 x (1/2)^8 x (1/2)^0 = 1 x (1/2)^8 = 1/256
Therefore, the probability that he answers at least 6 questions correctly is:
28/256 + 8/256 + 1/256 = 37/256
Answer: B
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