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I'm currently stuck on this question: A multiple choice test consists of 8 questions each with 4 choices. A student guesses the answers by choosing an answer at random. We can't use binomcdf or anything like that to find the answer

a) Compute the probability that the student answers 2 or more correctly.

b) If the student is able to eliminate one answer in each question as incorrect and chooses randomly from the other three choices, what is the probability the student answers 2 or more correctly?

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The simplest way to do (a) is to find the probability of answering 0 or only 1 correctly, then subtracting that from 1. With four possible answer to each problem, the probability of answering one correctly by choosing at random is 1/4 and the probability of answering incorrectly is 3/4. So the probability of answering all 8 questions incorrectly (0 correctly) is $(3/4)^8= 0.31640625$. The probability of answering a specific question correctly and the other seven incorrectly is $(1/4)(3/4)^7= 0.033370971679685$. The probability any one of the seven is correct and the other 3 incorrect is that times 8, 0.2669677734375. The probability that either all 8 are incorrect or one is correct and the other 7 are incorrect is 0.31640625+ 0.2669677734375= 0.5833740234375. The probability of answer 2 or more correctly is 1- 0.5833740234375= 0.4166259765625.

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