1: A given mid-exam contains 10 multiple choice questions, and each question has
four alternatives with one exact answer. Find the probability that the student exactly answered
A.3 questions
B.8 questions
C.At least 3 questions
D. Compute the mean and the variance of X.
To find the probability that the student exactly answered 3 questions correctly, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- n is the number of trials (10 multiple choice questions)
- k is the number of successful outcomes (3 correct answers)
- p is the probability of success on each trial (1/4, since there is one correct answer out of four alternatives)
Using this formula, we can calculate the probability:
P(X = 3) = (10 choose 3) * (1/4)^3 * (3/4)^(10-3)
P(X = 3) = (10! / (3! * (10-3)!)) * (1/4)^3 * (3/4)^7
= (10! / (3! * 7!)) * (1/4)^3 * (3/4)^7
= (10 * 9 * 8) / (3 * 2 * 1) * (1/64) * (2187/16384)
= (720 / 6) * (1/64) * (2187/16384)
= 120 * 1/64 * 0.13343048
= 0.26686096
Therefore, the probability that the student exactly answered 3 questions correctly is approximately 0.2669.
For part B, to find the probability that the student exactly answered 8 questions correctly, we use the same formula:
P(X = 8) = (10 choose 8) * (1/4)^8 * (3/4)^(10-8)
P(X = 8) = (10! / (8! * (10-8)!)) * (1/4)^8 * (3/4)^2
= (10! / (8! * 2!)) * (1/4)^8 * (3/4)^2
= (10 * 9) / (2 * 1) * (1/65536) * (9/16)
= 45/65536 * 9/16
= 0.00136757
Therefore, the probability that the student exactly answered 8 questions correctly is approximately 0.0014.
For part C, to find the probability that the student answered at least 3 questions correctly, we need to calculate the probability of answering 3, 4, 5, 6, 7, 8, 9, or 10 questions correctly and sum them up:
P(X >= 3) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
Using the same formula as before, we can calculate each probability and sum them:
P(X >= 3) = 0.26686096 + (10 choose 4) * (1/4)^4 * (3/4)^6 + (10 choose 5) * (1/4)^5 * (3/4)^5 + (10 choose 6) * (1/4)^6 * (3/4)^4 + (10 choose 7) * (1/4)^7 * (3/4)^3 + 0.00136757 + P(X = 9) + P(X = 10)
Using these calculations, we find that P(X >= 3) is approximately 0.9982.
For part D, we can calculate the mean and variance of X using the formulas:
Mean (μ) = n * p
Variance (σ^2) = n * p * (1 - p)
Mean (μ) = 10 * (1/4) = 2.5
Variance (σ^2) = 10 * (1/4) * (1 - 1/4) = 1.875
Therefore, the mean of X is 2.5 and the variance is 1.875.