1,assume test consists of 10 multiple choice questions with having 4 answer possibilities if students randomly guesses all questions then,
a, what is the probability that the gets four correct anwers?
b, what is the probability that student will be wrong in all questions?
2,suppose a manufacturer of tv tube draws a random sample of 10 tubes. the probability that a single tube selected at random defective is 0.1 then find.
a, th probability of finding no defective
b, the probability of exactly 3 defective
3, the probability of at most 2 defective
1a. To solve this problem, we need to use the binomial distribution. The probability of getting a correct answer by guessing is 1/4, and the probability of getting a wrong answer is 3/4. Let X be the number of correct answers. Then, X follows a binomial distribution with n = 10 and p = 1/4. The probability of getting exactly 4 correct answers is:
P(X = 4) = (10 choose 4) * (1/4)^4 * (3/4)^6 = 0.0000261
Therefore, the probability that the student gets four correct answers is 0.0000261.
1b. Again, we need to use the binomial distribution. The probability of getting all wrong answers is:
P(X = 0) = (10 choose 0) * (1/4)^0 * (3/4)^10 = 0.0563
Therefore, the probability that the student is wrong in all questions is 0.0563.
2a. The probability of finding no defective tube is:
P(X = 0) = (10 choose 0) * (0.9)^10 = 0.3487
Therefore, the probability of finding no defective tube is 0.3487.
2b. The probability of finding exactly 3 defective tubes is:
P(X = 3) = (10 choose 3) * (0.1)^3 * (0.9)^7 = 0.0574
Therefore, the probability of finding exactly 3 defective tubes is 0.0574.
3. The probability of at most 2 defective tubes is the sum of the probabilities of finding 0, 1, or 2 defective tubes:
P(X <= 2) = P(X = 0) + P(X = 1) + P(X = 2)
= (10 choose 0) * (0.9)^10 + (10 choose 1) * (0.1)^1 * (0.9)^9 + (10 choose 2) * (0.1)^2 * (0.9)^8
= 0.9298
Therefore, the probability of at most 2 defective tubes is 0.9298.
To answer the questions, we will use the concept of probability.
1. Problem:
Assume a test consists of 10 multiple-choice questions with 4 answer possibilities. If a student randomly guesses all the questions, what is the probability that:
a. The student gets four correct answers?
b. The student will be wrong in all questions?
Solution:
a. To find the probability of getting exactly four correct answers, we can use the binomial probability formula:
P(x) = C(n, x) * p^x * q^(n-x)
Here, n = number of trials or questions = 10,
x = the number of successes (in this case, correct answers) = 4,
p = probability of success (correct answer) = 1/4,
q = probability of failure (wrong answer) = 3/4.
Plugging these values into the formula:
P(4) = C(10, 4) * (1/4)^4 * (3/4)^(10-4)
Using the combination formula C(n, x) = n! / (x!(n-x)!):
P(4) = 10! / (4!(10-4)!) * (1/4)^4 * (3/4)^6
This can be simplified to:
P(4) = 210 * (1/256) * (729/4096)
P(4) ≈ 0.082
Therefore, the probability that the student gets exactly four correct answers is approximately 0.082, or 8.2%.
b. To find the probability that the student is wrong in all the questions, we need to calculate the probability of getting zero correct answers. Using the same formula as before:
P(0) = C(10, 0) * (1/4)^0 * (3/4)^(10-0)
Simplifying the expression:
P(0) = 1 * 1 * (19683/1048576)
P(0) ≈ 0.019
Therefore, the probability that the student is wrong in all questions is approximately 0.019, or 1.9%.
2. Problem:
A manufacturer of TV tubes draws a random sample of 10 tubes. The probability that a single tube selected at random is defective is 0.1. Find:
a. The probability of finding no defective tubes.
b. The probability of exactly 3 defective tubes.
Solution:
a. To find the probability of finding no defective tubes, we can use the binomial distribution formula, where the number of trials is 10 and there is a success (finding a defective tube) with probability 0.1. The probability of no defects is the probability of failure in all trials.
P(no defects) = (1 - p)^n
P(no defects) = (1 - 0.1)^10
P(no defects) = 0.9^10
P(no defects) ≈ 0.3487
Therefore, the probability of finding no defective tubes is approximately 0.3487, or 34.87%.
b. To find the probability of exactly 3 defective tubes, we use the binomial distribution formula again.
P(exactly 3 defects) = C(10, 3) * (0.1)^3 * (0.9)^(10-3)
P(exactly 3 defects) = 120 * 0.001 * 0.9^7
P(exactly 3 defects) ≈ 0.0574
Therefore, the probability of finding exactly 3 defective tubes is approximately 0.0574, or 5.74%.
3. Problem:
Find the probability of having at most 2 defective tubes in the same scenario.
Solution:
To find the probability of having at most 2 defective tubes, we need to calculate the sum of probabilities for 0, 1, and 2 defective tubes.
P(at most 2 defects) = P(0 defects) + P(1 defect) + P(2 defects)
Using the previous calculations:
P(at most 2 defects) ≈ 0.3487 + P(1 defect) + P(2 defects)
To find the individual probabilities for 1 and 2 defects, we can use the binomial distribution formula:
P(1 defect) = C(10, 1) * (0.1)^1 * (0.9)^(10-1)
P(1 defect) = 10 * 0.1 * 0.9^9
P(1 defect) ≈ 0.3874
P(2 defects) = C(10, 2) * (0.1)^2 * (0.9)^(10-2)
P(2 defects) = 45 * 0.01 * 0.9^8
P(2 defects) ≈ 0.1937
Substituting these values back into the equation:
P(at most 2 defects) ≈ 0.3487 + 0.3874 + 0.1937
P(at most 2 defects) ≈ 0.9298
Therefore, the probability of having at most 2 defective tubes is approximately 0.9298, or 92.98%.