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SC Moderator V
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A multiple-choice exam consists of 7 questions and each  [#permalink]

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1
9 00:00

Difficulty:   65% (hard)

Question Stats: 53% (02:05) correct 47% (02:01) wrong based on 57 sessions

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A multiple-choice exam consists of 7 questions and each question has 3 answer choices. For every question there is exactly one correct answer. If a student randomly guesses at each question, what is the probability that she will correctly answer exactly 4 questions?
A) $$\frac{8}{3^7}$$
B)$$\frac{1}{3^4}$$
C)$$\frac{240}{3^7}$$
D)$$\frac{280}{3^7}$$
E)$$\frac{140}{3^5}$$

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Senior Manager  P
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Re: A multiple-choice exam consists of 7 questions and each  [#permalink]

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4
1
Correct in 1 way. Not correct in two ways.
CCCCNNN
= 7!/4!3! = 35.
Since each not correct can have two possibilities ----> total favourable = 2*2*2*35 = 280
Total outcomes = 3*3*3*3*3*3*3 = 3^7
D
##### General Discussion
Intern  B
Joined: 19 Sep 2016
Posts: 45
Re: A multiple-choice exam consists of 7 questions and each  [#permalink]

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can someone explain in detail ??
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3410
Re: A multiple-choice exam consists of 7 questions and each  [#permalink]

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Solution

Given:
• A multiple-choice exam consists of 7 questions

To find:
• Probability that she will correctly answer exactly 4 questions.

Approach and Working:
Probability to correctly answer exactly 4 questions= $$\frac{Total\ ways\ to\ correctly\ answer\ 4\ questions}{Total\ ways\ to\ mark\ 7\ questions}$$.

Total ways to mark 7 questions:

• Each question can be marked in 3 ways.
o Either 1st option or 2nd or 3rd.
• Therefore, ways to answer 7 questions= 3*3*3*3*3*3*3= $$3^7$$.

Ways to mark 4 questions correctly

• Out of 7 questions, we need to select 4 questions whose answers must be correct.
o And, we can select the 4 correct questions in 7c4 ways.
o Now, for each correct question, there can only be one way- Correct option.
o However, for each incorrect question, there can be two ways- 2 incorrect options.
 Hence, total ways to mark 4 questions correct= 7c4* 1*2*2*2= 35*8= 280

Hence, Probability to correctly answer exactly 4 questions= $$\frac{280}{3^7}$$

Hence, the correct answer is option D.

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Manager  G
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Re: A multiple-choice exam consists of 7 questions and each  [#permalink]

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ShankSouljaBoi wrote:
Correct in 1 way. Not correct in two ways.
CCCCNNN
= 7!/4!3! = 35.
Since each not correct can have two possibilities ----> total favourable = 2*2*2*35 = 280
Total outcomes = 3*3*3*3*3*3*3 = 3^7
D

Great explanation, just to tweak a little

desired outcome CCCCWWW :
the probability for correct answer is $$\frac{1}{3}$$ and same for wrong answer is $$\frac{2}{3}$$
we can directly write P(CCCCWWW)= $$\frac{7!}{4!3!}*\frac{1}{3}*\frac{1}{3}*\frac{1}{3}*\frac{1}{3}*\frac{2}{3}*\frac{2}{3}*\frac{2}{3}$$ = $$\frac{280}{3^7}$$
Senior Manager  P
Joined: 27 Feb 2014
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GMAT 1: 570 Q49 V20 GPA: 3.97
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Re: A multiple-choice exam consists of 7 questions and each  [#permalink]

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Abhi077 wrote:
A multiple-choice exam consists of 7 questions and each question has 3 answer choices. For every question there is exactly one correct answer. If a student randomly guesses at each question, what is the probability that she will correctly answer exactly 4 questions?
A) $$\frac{8}{3^7}$$
B)$$\frac{1}{3^4}$$
C)$$\frac{240}{3^7}$$
D)$$\frac{280}{3^7}$$
E)$$\frac{140}{3^5}$$

Probability of getting exactly 4 correct and 3 incorrect is (1/3)^4 x (2/3)^3
Out of 7 questions selecting 4 correct questions is 7C4

required probability = 7C4 x (1/3)^4 x (2/3)^3
= 280/3^7

D is correct
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Inspired by great content in some best books on GMAT, I have created my own YouTube channel-QUANT MADE EASY! I would love some support and feedback. Please hit subscribe and check it out! Re: A multiple-choice exam consists of 7 questions and each   [#permalink] 26 Nov 2019, 01:52

# A multiple-choice exam consists of 7 questions and each  