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A person is taking a 77​-question ​multiple-choice test for which each question has three answer​ choices, only one of which is correct. The person decides on answers by rolling a fair die and marking the first answer choice if the die shows 1 or​ 2, the second if it shows 3 or​ 4, and the third if it shows 5 or 6. Find the probability that the person gets fewer than 33 correct answersanswers?

Answer :

jennerfranco

Answer:

0.9488

Step-by-step explanation:

We have a binomial experiment with n = 77 independent trials, in each trial, we can have a success or a fail, the probability of success is p = 1/3 (q = 2/3) because each question has three answer choices and only one is correct, besides the person decides on answers by rolling a fair die. Therefore, the probability that the person gets exactly x correct answers is given by  

[tex]P(X = x) = (77Cx)p^{x}q^{n-x}[/tex] and the probability that the person gets fewer than 33 correct answers is given by [tex]P(X < 33) = \sum_{x=1}^{32}(77Cx)(1/3)^{x}(2/3)^{77-x}[/tex] = 0.9488

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