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On a multiple choice test, there are four answer choices for each

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Joined: 20 Feb 2017
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On a multiple choice test, there are four answer choices for each  [#permalink]

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New post 26 Aug 2018, 19:40
6
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

64% (02:03) correct 36% (01:45) wrong based on 47 sessions

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On a multiple choice test, there are four answer choices for each question, and for each question there is exactly one correct answer. If the test consists of four questions, and a student guesses randomly at each question, what is the probability the student will get either no correct answers or exactly one correct answer?

A)5/4^4

B)27/4^4

C)65/4^4

D)108/4^4

E)189/4^4

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Re: On a multiple choice test, there are four answer choices for each  [#permalink]

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New post 26 Aug 2018, 20:03
Raksat wrote:
On a multiple choice test, there are four answer choices for each question, and for each question there is exactly one correct answer. If the test consists of four questions, and a student guesses randomly at each question, what is the probability the student will get either no correct answers or exactly one correct answer?

A)5/4^4

B)27/4^4

C)65/4^4

D)108/4^4

E)189/4^4


P(0 or 1) = P(0) + P(1) = (3/4)^4 + 4*(3/4)^3*1/4 = 189/4^4.

Answer: E.
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Re: On a multiple choice test, there are four answer choices for each  [#permalink]

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New post 27 Aug 2018, 08:35
Raksat wrote:
On a multiple choice test, there are four answer choices for each question, and for each question there is exactly one correct answer. If the test consists of four questions, and a student guesses randomly at each question, what is the probability the student will get either no correct answers or exactly one correct answer?

A)5/4^4

B)27/4^4

C)65/4^4

D)108/4^4

E)189/4^4



no of ways he can get all incorrect answers = 1
p(all incorrect) = 1*3^4/4^4

no of ways he can get exactly 1 correct = 4
p(one correct) = 4*3^3/4^4

p(all incorrect) + p(exactly one correct) = 3^4/4^4 + 4*3^3/4^4
=189/4^4
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Re: On a multiple choice test, there are four answer choices for each  [#permalink]

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New post 27 Aug 2018, 13:31
P(Getting all incorrect or getting one correct) = P(Getting all incorrect) + P(getting one correct) ---- > (1)
P(Getting all incorrect) = P(getting incorrect 1st question) * P(getting incorrect 2nd question) * P(getting incorrect 3rd question) * P(getting incorrect 4th question)
= \(3/4 * 3/4 * 3/4 * 3/4 = (3/4)^4\) --------> (2)
P(Getting one correct) = [ P(getting 1st question correct ) * P(getting incorrect 2nd question) * P(getting incorrect 3rd question) * P(getting incorrect 4th question) * 4 ]
= \((1/4 * 3/4 * 3/4 * 3/4) * 4 = (3/4)^3\) --------> (3)

Substituting (2) and (3) in (1)
= \(((3/4)^3(3/4 +1)\)
= \(189/4^4\)

Answer: E
GMAT Club Bot
Re: On a multiple choice test, there are four answer choices for each &nbs [#permalink] 27 Aug 2018, 13:31
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