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A multiple choice examination has 5 questions

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prashanths
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A multiple choice examination has 5 questions [#permalink] New post   17 Jun 2019, 02:04
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A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more answers correct just by guessing is

A) 17/(3^5)
B) 13/(3^5)
C) 11/(3^5)
D) 10/(3^5)
E) 12/(3^5)
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C
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chetan2u
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Re: A multiple choice examination has 5 questions [#permalink] New post   17 Jun 2019, 05:18
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prashanths wrote:
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more answers correct just by guessing is

A) 17/(3^5)
B) 13/(3^5)
C) 11/(3^5)
D) 10/(3^5)
E) 12/(3^5)


Total ways in which the questions can be answered...\(3*3*3*3*3=3^5\)

Ways atleast 4 are correct..
1) exactly 4 are correct..
Choose the 4 out of 5 which can be answered in just one way and the 5ty can be answered in 2 ways(2 wrong answers)=5C4*2=10
2) all 5 correct..
5C5=1
Total 10+1=11

Probability of atleast 4 correct= \(\frac{11}{3^5}\)


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Re: A multiple choice examination has 5 questions [#permalink] New post   17 Jun 2019, 07:52
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The probability to get exactly 4 questions correct= \(5C4*(\frac{1}{3})^4*(\frac{2}{3})\)=\(\frac{10}{3^5}\)
The probability to get exactly 5 questions correct=\(5C5*(\frac{1}{3})^5\)= \(\frac{1}{3^5}\)

The probability that a student will get 4 or more answers correct just bu guessing= \(\frac{10}{3^5}+\frac{1}{3^5}\)=\(\frac{11}{3^5}\)

prashanths wrote:
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more answers correct just by guessing is

A) 17/(3^5)
B) 13/(3^5)
C) 11/(3^5)
D) 10/(3^5)
E) 12/(3^5)
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Lodz697
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Re: A multiple choice examination has 5 questions [#permalink] New post   04 Mar 2024, 04:22
Why can't I use a "probability way":

Four correct: (1/3)^4 x (2/3) = 2/243

Five correct: (1/3)^5 = 1/243

2/243 + 1/243 = 3/243 (which is obv wrong but I don't understand why).

Bunuel
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Bunuel
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Re: A multiple choice examination has 5 questions [#permalink] New post   04 Mar 2024, 05:02
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Lodz697 wrote:
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more answers correct just by guessing is

A) 17/(3^5)
B) 13/(3^5)
C) 11/(3^5)
D) 10/(3^5)
E) 12/(3^5)


Why can't I use a "probability way":

Four correct: (1/3)^4 x (2/3) = 2/243

Five correct: (1/3)^5 = 1/243

2/243 + 1/243 = 3/243 (which is obv wrong but I don't understand why).

Bunuel

It should be:

P(4 correct) + P(5 correct) =

= (1/3)^4 *2/3*5!/4! +  (1/3)^5 = 

= 11/243.

We multiply (1/3)^4 *2/3 by 5!/4! = 5 becasue CCCCW can be obtained in different ways: CCCCW, CCCWC, CCWCC, CWCCC, and WCCCC.

Answer: C.
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Re: A multiple choice examination has 5 questions [#permalink]
04 Mar 2024, 05:02
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