Answer:
a. 5.74%.
b. 7.29%
c. 20.18%
d. 34.87%
e. 65.13%
Explanation:
a. This problem can be solved using the binomial distribution formula: P(X=k) = (n choose k) * p^k * (1-p)^(n-k), where n is the sample size, k is the number of successes, p is the probability of success, and (n choose k) is the binomial coefficient.
For this problem, n=7, p=0.1, and we want to find P(X=3). Therefore, we have:
P(X=3) = (7 choose 3) * 0.1^3 * (0.9)^4 = 0.0574, or 5.74%.
b. We have n=5, p=0.1, and we want to find P(X=2). Therefore, we have:
P(X=2) = (5 choose 2) * 0.1^2 * (0.9)^3 = 0.0729, or 7.29%.
c. To find the probability that 4 or 5 out of 10 bulbs are defective, we can use the binomial distribution to find the probabilities of each outcome separately and add them together. We have n=10 and p=0.1.
P(4 out of 10 are defective) = (10 choose 4) * 0.1^4 * (0.9)^6 = 0.1937, or 19.37%.
P(5 out of 10 are defective) = (10 choose 5) * 0.1^5 * (0.9)^5 = 0.0081, or 0.81%.
P(4 or 5 out of 10 are defective) = P(4 out of 10 are defective) + P(5 out of 10 are defective) = 0.1937 + 0.0081 = 0.2018, or 20.18%.
d. To find the probability that no bulbs out of 10 are defective, we can use the binomial distribution with n=10 and p=0.1, and find P(X=0). Therefore, we have:
P(X=0) = (10 choose 0) * 0.1^0 * (0.9)^10 = 0.3487, or 34.87%.
e. To find the probability that one or more bulbs out of 10 are defective, we can use the complement rule and subtract the probability of no bulbs being defective from 1. Therefore, we have:
P(one or more bulbs out of 10 are defective) = 1 - P(X=0) = 1 - 0.3487 = 0.6513, or 65.13%.