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A multiple-choice exam consists of 7 questions and each

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

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New post 14 Oct 2018, 00:03
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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)\(\frac{140}{3^5}\)
Spoiler: OA

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

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New post 14 Oct 2018, 00:22
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
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Re: A multiple-choice exam consists of 7 questions and each  [#permalink]

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New post 14 Oct 2018, 13:25
can someone explain in detail ??
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Re: A multiple-choice exam consists of 7 questions and each  [#permalink]

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New post 14 Oct 2018, 16:07

Solution


Given:
    • A multiple-choice exam consists of 7 questions
    • Each question has 3 answer choices and one correct answer.

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.

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

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New post 15 Oct 2018, 07:46
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)= \(\frac{7!}{4!3!}*\frac{1}{3}*\frac{1}{3}*\frac{1}{3}*\frac{1}{3}*\frac{2}{3}*\frac{2}{3}*\frac{2}{3}\) = \(\frac{280}{3^7}\)
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Re: A multiple-choice exam consists of 7 questions and each  [#permalink]

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New post 26 Nov 2019, 01:52
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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Re: A multiple-choice exam consists of 7 questions and each   [#permalink] 26 Nov 2019, 01:52
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