Question

A person is taking a 13​-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 3 correct answer

Answer 1

The probability of getting fewer than 3 correct answers is 0.1385

How calculate the probability of getting fewer than 3 correct answers?

This is a binomial problem with n = 13

Since we are dealing with a binomial probability distribution. We are going to use the binomial distribution formula for determining the probability of x successes:

P(x = r) = nCr . p^r  . q^n-r

total choices = 13 x 3 = 39

p = P(correct answer) = 13/39 =1/3

Given: n = 13, p = 1/3, q = 1- 1/3 = 2/3

Fewer than 3 correct answers mean less than 3 correct answers

P(x<3) = P(x = 0) + P(x=1) + P(x=2)

P(x<3) = (13C0 × (1/3)^0 × (2/3)^13-0) + (13C1 × (1/3)^1 × (2/3)^13-1 + (13C2 × (1/3)^2 × (2/3)^13-2)

         = 0.005138 + 0.0333 + 0.1001

         = 0.1385

Therefore, the probability that the person gets fewer than 3 correct answers is 0.1385

Learn more about binomial distribution on:

brainly.com/question/29220138

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