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May 29th, 2017, 07:22 PM   #1
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Probability

Each question in a test has 4 options of which only one is correct. Ashok does not know which of the options are correct or wrong in 3 questions. He decides to select randomly the options for these 3 questions independently. The probability that he will choose at least 2 correctly is
A) more than 0.25
B) in the interval (0.2,0.25]
C) in the interval (1/6,0.2]
D) less than 1/6

Please help me to solve this problem and let me know how can we solve similar kind of these issues.


Thank you.

Last edited by skipjack; May 30th, 2017 at 12:17 AM.
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May 29th, 2017, 09:40 PM   #2
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the probability of guessing $k$ correct out of $3$ is described by a binomial distribution with parameters $p=\dfrac 1 2,~n=3$

$P[k]=\dbinom{3}{k}\left(\dfrac 1 2\right)^3$

He gets 1 right because he knows the correct answer. So getting 2 or more right is guessing 1 or more correctly.

$P[k\geq 1] = 1 - P[k=0] = 1 - \dfrac{\binom{3}{0}}{8} = \dfrac 7 8$

so

$P[\text{2 or more correct}] = 0.875$
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May 30th, 2017, 01:09 AM   #3
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That's incorrect, as it misreads the description and uses an incorrect value for $p$.

The probability of choosing exactly 2 correctly is $\binom{3}{2}(1/4)^2(3/4)^1 = 9/64$.
The probability of choosing exactly 3 correctly is $\binom{3}{3}(1/4)^3(3/4)^0 = 1/64$.
The probability of choosing at least 2 correctly is therefore 10/64, which is less than 10/60, so (D) is the answer.
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