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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] New post 07 Aug 2012, 05:58
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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
[Reveal] Spoiler: OA

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

Answer: B.

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] New post 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}.

Answer B

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

So total probability
=106/1024

=53/512
Ans B
Re: In a multiple choice test comprising 5 Questions, each with   [#permalink] 07 Aug 2012, 07:16
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