Final answer:
To calculate the probabilities in this scenario, we can use the concept of binomial distribution. The probability that no samples are mutated is approximately 0.8687. The probability that at most one sample is mutated is approximately 0.99.
Step-by-step explanation:
To determine the probabilities in this scenario, we can use the concept of binomial distribution. Let's break down each part:
(a) No samples are mutated:
This means that all 14 samples are not mutated. The probability of a sample being mutated is 1%, so the probability of a sample not being mutated is 99%. Since the samples are independent, we can use the binomial distribution formula: P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of samples, k is the number of mutated samples, and p is the probability of a sample being mutated.
Plugging in the values: P(X=0) = C(14,0) * (0.01)^0 * (0.99)^(14-0)
Simplifying:
P(X=0) = 1 * 1 * (0.99)^14
Calculating:
P(X=0) ≈ 0.8687
So, the probability that no samples are mutated is approximately 0.8687.
(b) At most one sample is mutated:
This means that either no samples are mutated or only one sample is mutated. We can calculate this probability by summing the individual probabilities of these two cases: P(X=0) + P(X=1)
Using the values we calculated in part (a): 0.8687 + P(X=1)
To calculate P(X=1), we can use the binomial distribution formula again: P(X=1) = C(14,1) * (0.01)^1 * (0.99)^(14-1)
Calculating:
P(X=1) ≈ 0.1213
So, the probability that at most one sample is mutated is approximately 0.8687 + 0.1213 = 0.99.
(c) More than half the samples are mutated:
This means that 8 or more samples are mutated. We can calculate this probability by summing the individual probabilities of these cases: P(X=8) + P(X=9) + ... + P(X=14)
Using the binomial distribution formula again, we can calculate each individual probability:
P(X=k) = C(14,k) * (0.01)^k * (0.99)^(14-k), where k ranges from 8 to 14.
Calculating each probability and summing them:
P(X=8) + P(X=9) + ... + P(X=14) ≈ 0.0002 + 0.0004 + 0.0009 + 0.0018 + 0.0035 + 0.0066 + 0.0119 ≈ 0.0253
So, the probability that more than half the samples are mutated is approximately 0.0253.