On a six-question multiple-choice test there are five possible answers for each question, of which one is correct (C) and four are incorrect (I). If a student guesses randomly and independently, find the probability of (a) Being correct only on questions 1 and 4 (i. e., scoring C, I, I, C, I, I). (b) Being correct on two questions.

(a) 1.64%

(b) 24.58%

Step-by-step explanation:

As for each question there are one correct answer in five possible answers, the probability of guessing the correct answer is 1/5 = 0.2, so the probability of guessing the wrong answer is (1 - 0.2) = 0.8.

(a) The probability of being correct only on questions 1 and 4 is calculated multiplying all the following probabilities:

question 1 correct: 0.2

question 2 wrong: 0.8

question 3 wrong: 0.8

question 4 correct: 0.2

question 5 wrong: 0.8

question 6 wrong: 0.8

P = (0.2)^2 * (0.8)^4 = 0.0164 = 1.64%

(b) The probability of being correct in two questions is calculated similarly to the question in (a), but now we also have a combination problem: The 2 correct questions can be any group of 2 questions inside the 6 total questions, so we also multiply the probability of each question being correct or wrong by the combination of 6 choose 2:

C(6,2) = 6!/(4!*2!) = 6*5/2 = 15

P = 15 * (0.2)^2 * (0.8)^4 = 0.2458 = 24.58%

the absolute value function

my answer was deleted because it was to vague the answer is 12.76 becuase its in the hundreths place. john