A student attempts a test with 10 multiple-choice questions, where each question has 4 possible options. She does not know the correct answer to any of the questions and randomly selects one of the 4 options for each of them. What is the probability that she will get 3 questions correct?10C3(1/4)1010C3(3/4)3(1/4)710C3(1/4)3(3/4)7Binomial probability cannot be used here.
Question
A student attempts a test with 10 multiple-choice questions, where each question has 4 possible options. She does not know the correct answer to any of the questions and randomly selects one of the 4 options for each of them. What is the probability that she will get 3 questions correct?10C3(1/4)1010C3(3/4)3(1/4)710C3(1/4)3(3/4)7Binomial probability cannot be used here.
Solution
The problem can be solved using the binomial probability formula, which is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the number of combinations of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
In this case:
- n = 10 (the number of questions)
- k = 3 (the number of correct answers)
- p = 1/4 (the probability of guessing a question correctly, since there are 4 options)
So, the probability that the student will get 3 questions correct is:
P(X=3) = C(10, 3) * ((1/4)^3) * ((1 - 1/4)^(10 - 3))
Calculating this gives:
P(X=3) = 120 * (1/64) * (3/4)^7 ≈ 0.2503
So, the probability that the student will get exactly 3 questions correct by guessing is approximately 0.2503, or 25.03%.
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