In a multiple choice exam with 10 questions, each question has 4 options.
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In a multiple choice exam with 10 questions, each question has 4 options. A student needs to get at least 3 questions correct in order to pass the exam. Tina has not studied for the exam and decides to mark the options randomly. What is the probability that she passes the exam?
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Tina passes if she gets 3 or more correct. Using the binomial distribution with n=10, p=1/4 (guessing correctly), the probability of failing is P(X≤2) = P(X=0) + P(X=1) + P(X=2) = C(10,0)(1/4)⁰(3/4)¹⁰ + C(10,1)(1/4)¹(3/4)⁹ + C(10,2)(1/4)²(3/4)⁸ = (3/4)¹⁰ + 10(1/4)(3/4)⁹ + 45(1/16)(3/4)⁸ = (3/4)⁸[(3/4)² + 10(1/4)(3/4) + 45(1/16)] = (3/4)⁸[(9/16) + (30/16) + (45/16)] = (84/16)(3/4)⁸ = (21/4)(3/4)⁸. So P(pass) = 1 - (21/4)(3/4)⁸.
Step-by-step Derivation:
Step 1: Identify the model. Random guessing on 4-option MCQs means P(correct) = 1/4, P(wrong) = 3/4 per question.
Step 2: Pass requires getting ≥3 correct out of 10, so fail requires ≤2 correct.
Step 3: Calculate P(X ≤ 2) using binomial distribution:
P(X=k) = C(n,k) × p^k × (1-p)^(n-k)
P(X=0) = C(10,0) × (1/4)⁰ × (3/4)¹⁰ = (3/4)¹⁰
P(X=1) = C(10,1) × (1/4)¹ × (3/4)⁹ = 10 × (1/4) × (3/4)⁹
P(X=2) = C(10,2) × (1/4)² × (3/4)⁸ = 45 × (1/16) × (3/4)⁸
Step 4: Factor out (3/4)⁸:
P(fail) = (3/4)⁸ × [(3/4)² + 10×(1/4)×(3/4) + 45×(1/16)]
= (3/4)⁸ × [(9/16) + (30/16) + (45/16)]
= (3/4)⁸ × (84/16)
= (3/4)⁸ × (21/4)
= (21/4) × (3/4)⁸
Step 5: P(pass) = 1 - P(fail) = 1 - (21/4) × (3/4)⁸