Question

On a multiple choice examination, each question has exactly five options, of which the student must pick one. Only one option is correct for each question. If the student is merely guessing at the answer, each option should be equally likely to be chosen. Given that the test has 18 questions, and that a student is just guessing on each question, find the probability of the following events: (a) exactly three are correct, (b) fewer than five are correct, (c) between two and sever answers are correct.

Answer 1

Please note ⁿC_x = n! / [(n-x)!*x!]

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Here n = 18, p = 0.2, q = 1 – p = 0.8.

___________________________________________

(a) P(X = 3) = ¹⁸C₃ * (0.2)³ * (0.8)^18-3 = 0.2297

___________________________________________

(b) P(Fewer than 5) = P(X < 5) = P(0) + P(1) + P(2) + P(3) + P(4)

P(X = 0) = ¹⁸C₀ * (0.2)⁰ * (0.8)^18-0 = 0.018

P(X = 1) = ¹⁸C₁ * (0.2)¹ * (0.8)^18-1 = 0.0811

P(X = 2) = ¹⁸C₂ * (0.2)² * (0.8)^18-2 = 0.1723

P(X = 3) = ¹⁸C₃ * (0.2)³ * (0.8)^18-3 = 0.2297

P(X = 4) = ¹⁸C₄ * (0.2)⁴ * (0.8)^18-4 = 0.2153

Therefore P(Fewer than 5) = 0.018 + 0.0811 + 0.1723 + 0.2297 + 0.2153 = 0.7164

___________________________________________

(c) P(Between 2 and 7 are correct) = P(3) + P(4) + P(5) + P(6)

P(X = 3) = ¹⁸C₃ * (0.2)³ * (0.8)^18-3 = 0.2297

P(X = 4) = ¹⁸C₄ * (0.2)⁴ * (0.8)^18-4 = 0.2153

P(X = 5) = ¹⁸C₅ * (0.2)⁵ * (0.8)^18-5 = 0.1507

P(X = 6) = ¹⁸C₆ * (0.2)⁶ * (0.8)^18-6 = 0.0816

Therefore P(Between 2 and 7) = 0.2297 + 0.2153 + 0.1507 + 0.0816 = 0.6773

___________________________________________