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 May 29th, 2017, 07:22 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 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.
 May 29th, 2017, 09:40 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,931 Thanks: 999 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 $P[\text{2 or more correct}] = 0.875$
 May 30th, 2017, 01:09 AM #3 Global Moderator   Joined: Dec 2006 Posts: 18,954 Thanks: 1601 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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