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.
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
1 answer