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17 Jun 2019, 02:04
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (02:00) correct
40%
(02:29)
wrong
based on 128
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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)
A) 17/(3^5)
B) 13/(3^5)
C) 11/(3^5)
D) 10/(3^5)
E) 12/(3^5)
17 Jun 2019, 05:18
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prashanths
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= \(\frac{11}{3^5}\)
C
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= \(\frac{10}{3^5}+\frac{1}{3^5}\)=\(\frac{11}{3^5}\)
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
