Question

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.

Answer 1

The probability of being correct on questions 1 and 4 only on a six-question test with five answer choices is (1/5)^2 × (4/5)^4, while the probability of being correct on exactly two questions is 15 × (1/5)^2 × (4/5)^4.

Step-by-step explanation:

The probability of a student being correct only on questions 1 and 4 (C, I, I, C, I, I) on a multiple-choice test with five possible answers for each question involves calculating the probability for each outcome and multiplying them together. For being correct (C), the probability is 1/5, and for being incorrect (I), it is 4/5. Therefore:

Probability(C, I, I, C, I, I) = (1/5) × (4/5) × (4/5) × (1/5) × (4/5) × (4/5)

This simplifies to:

Probability(C, I, I, C, I, I) = (1/5)^2 × (4/5)^4

For part (b), the probability of being correct on exactly two questions means any two questions could be correct and the others incorrect. We can use the combination formula to determine the number of ways to choose 2 correct answers out of 6, and then multiply by the probability raised to the appropriate powers:

Number of ways to choose 2 out of 6 = 6 choose 2 = 6! / (2!(6-2)!) = 15

Probability of 2 correct answers = 15 × (1/5)^2 × (4/5)^4

Answer 2

(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%