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# In a multiple choice test comprising 5 Questions, each with

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In a multiple choice test comprising 5 Questions, each with  [#permalink]

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07 Aug 2012, 05:58
3
5
00:00

Difficulty:

85% (hard)

Question Stats:

63% (02:40) correct 37% (02:51) wrong based on 138 sessions

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In a multiple choice test comprising 5 Questions, each with 4 choices, what is the probability of a student getting 3 or more questions correct? Each question has only one correct answer and the student is equally likely to choose any of the four choices.

A) 24/256
B) 53/512
C) 105/512
D) 459/512
E) 47/256

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Re: In a multiple choice test comprising 5 Questions, each with  [#permalink]

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07 Aug 2012, 07:02
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SOURH7WK wrote:
In a multiple choice test comprising 5 Questions, each with 4 choices, what is the probability of a student getting 3 or more questions correct? Each question has only one correct answer and the student is equally likely to choose any of the four choices.

A) 24/256
B) 53/512
C) 105/512
D) 459/512
E) 47/256

The probability of getting correct (C) answer is 1/4 and the the probability of getting wrong (W) answer is 3/4.

We need to find the probability of getting at 3, 4 or 5 correct answers:

The probability of getting 3 correct answers out of 5 is $$\frac{5!}{3!2!}*(\frac{1}{4})^3*(\frac{3}{4})^2$$, we are multiplying by $$\frac{5!}{3!2!}$$ since CCCWW scenario can occur in several ways: CCCWW, WWCCC, CWCWC, ... basically the number of permutations of 5 letters CCCWW, out of which 3 C's and 2 W's are identical ($$\frac{5!}{3!2!}$$);

The probability of getting 4 correct answers out of 5 is $$\frac{5!}{4!}*(\frac{1}{4})^4*(\frac{3}{4})$$;

The probability of getting 5 correct answers out of 5 is simply $$(\frac{1}{4})^5$$.

The overall probabity will be the sum of the above probabilities: $$\frac{5!}{3!2!}*(\frac{1}{4})^3*(\frac{3}{4})^2+\frac{5!}{4!}*(\frac{1}{4})^4*(\frac{3}{4})+(\frac{1}{4})^5=\frac{53}{512}$$.

P.S. Not a GMAT type of qustion.
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Re: In a multiple choice test comprising 5 Qs, each with 4 choic  [#permalink]

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07 Aug 2012, 07:06
SOURH7WK wrote:
In a multiple choice test comprising 5 Questions, each with 4 choices, what is the probability of a student getting 3 or more questions correct? Each question has only one correct answer and the student is equally likely to choose any of the four choices.

A) 24/256
B) 53/512
C) 105/512
D) 459/512
E) 47/256

We can compute either the probability of answering 3, 4, or 5 questions correctly, or subtract from 1 the probability of answering 0, 1, or 2 questions.
The probability of answering any question correctly is 1/4, and obviously, answering incorrectly is 3/4.

P(3 correct answers) =$$\frac{5*4}{2}(\frac{1}{4})^3(\frac{3}{4})^2=\frac{90}{4^5}$$
P(4 correct answers) =$$5(\frac{1}{4})^4(\frac{3}{4})=\frac{15}{4^5}$$
P(5 correct answers) =$$(\frac{1}{4})^5=\frac{1}{4^5}$$
The sum of the above probabilities gives $$\frac{106}{4^5}=\frac{53}{512}$$.

Relevant topic for this question - Binomial probability distribution
Being a straightforward technical question, I question its importance for a real test.
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Re: In a multiple choice test comprising 5 Questions, each with  [#permalink]

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07 Aug 2012, 07:16
Here is my approach-
Probability for choosing correct answer is 1/4 and incorrect answer is 3/4
C(5,3)*(1/4)^3*(3/4)^2=90/(4^5)
C(5,4)*(1/4)^4*(3/4)^1=15/(4^5)
C(5,5)*(1/4)^5*(3/4)^0=1/(4^5)

So total probability
=106/1024

=53/512
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Re: In a multiple choice test comprising 5 Questions, each with  [#permalink]

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18 Mar 2016, 18:59
SOURH7WK wrote:
In a multiple choice test comprising 5 Questions, each with 4 choices, what is the probability of a student getting 3 or more questions correct? Each question has only one correct answer and the student is equally likely to choose any of the four choices.

A) 24/256
B) 53/512
C) 105/512
D) 459/512
E) 47/256

we need to find the probability when:
3 questions are answered correctly, 2 incorrectly
4 correctly, 1 incorrectly
and 5 correctly.

1 case: (1/4)^3 * (3/4)^2 = 9/1024. since we have 5 questions, and we need to select 3 out of these, we can answer correctly 3 out of 5 in 5C3 ways. or total probability 90/1024
2 case: (1/4)^3 * (3/4) = 3/1024. we can answer 4 correctly in 5C4 ways. total probability 15/1024
3 case: (1/4)^5. we can answer all 5 correctly in 1 way. so 1/1024

total: 90+15+1/1024 = 106/1024 = 53/512

B
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Posts: 40
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GMAT 1: 720 Q49 V40
GPA: 4
WE: Engineering (Energy and Utilities)
Re: In a multiple choice test comprising 5 Questions, each with  [#permalink]

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31 Oct 2018, 06:26
Bunuel wrote:
SOURH7WK wrote:
In a multiple choice test comprising 5 Questions, each with 4 choices, what is the probability of a student getting 3 or more questions correct? Each question has only one correct answer and the student is equally likely to choose any of the four choices.

A) 24/256
B) 53/512
C) 105/512
D) 459/512
E) 47/256

The probability of getting correct (C) answer is 1/4 and the the probability of getting wrong (W) answer is 3/4.

We need to find the probability of getting at 3, 4 or 5 correct answers:

The probability of getting 3 correct answers out of 5 is $$\frac{5!}{3!2!}*(\frac{1}{4})^3*(\frac{3}{4})^2$$, we are multiplying by $$\frac{5!}{3!2!}$$ since CCCWW scenario can occur in several ways: CCCWW, WWCCC, CWCWC, ... basically the number of permutations of 5 letters CCCWW, out of which 3 C's and 2 W's are identical ($$\frac{5!}{3!2!}$$);

The probability of getting 4 correct answers out of 5 is $$\frac{5!}{4!}*(\frac{1}{4})^4*(\frac{3}{4})$$;

The probability of getting 5 correct answers out of 5 is simply $$(\frac{1}{4})^5$$.

The overall probabity will be the sum of the above probabilities: $$\frac{5!}{3!2!}*(\frac{1}{4})^3*(\frac{3}{4})^2+\frac{5!}{4!}*(\frac{1}{4})^4*(\frac{3}{4})+(\frac{1}{4})^5=\frac{53}{512}$$.

P.S. Not a GMAT type of qustion.

Hi,

Can we not solve the question by calculating the probability of minimum 2 wrongs out of 5 questions?
P (at least of 3 correct) = 1 - P(minimum 2 wrongs)
Re: In a multiple choice test comprising 5 Questions, each with &nbs [#permalink] 31 Oct 2018, 06:26
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