Question

On a multiple-choice exam with four possible answers for each of the five questions, what is the probability that a student would get four or more correct answers just by guessing? Hint: Use the fact that P(E∩F)=P(E)⋅P(F) for two independent event (generalized for more than events). Getting one answer correct is independent of another. Also P( at least 4)=P( exactly 4)+P( exactly 5), the last events are mutually exclusive so P(A∪B)= P(A)+P(B)

Choose the all of the R-codes that compute the desired probability, could be more than one correct answer:

dbinom(x=4, size =5, prob =1/4)+dbinom(x=5, size =5, prob =1/4)
pbinom(q=3, size =5, prob =1/4, lower.tail = FALSE )
choose (5,4) ∗ (1/4) ∧4 ∗(3/4) ∧1+choose(5,5) ∗(1/4)∧5 ∗(2/4) ∧ 0
pbinom(q=3, size =5, prob =1/4, lower.tail = TRUE )

Answer 1

To find the probability of getting four or more correct answers just by guessing on a multiple-choice exam with four possible answers for each of the five questions, the probability can be calculated using the binomial distribution formula.

Step-by-step explanation:

To find the probability of getting four or more correct answers just by guessing on a multiple-choice exam with four possible answers for each of the five questions, we can use the formula P(A≥4)=P(A=4)+P(A=5), where A represents the number of correct answers. Each answer is independent, so the probability of guessing a correct answer is 1/4.

Therefore, the probability can be calculated as:

P(A≥4) = dbinom(x=4, size =5, prob =1/4) + dbinom(x=5, size =5, prob =1/4) = 0.0884 + 0.0026 = 0.0910.