The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of ten syringes taken from the batch. Suppose the batch contains 2% defective syringes. What is the probability that the batch will be accepted? Round your answer to three decimal places. my answer was o.984 which was incorrect, my possible answer trying this again was 0.016
0.984
0.016
0.999
0.817
0.000
To answer this question, we need to use the concept of binomial probability. In this case, we have a batch of syringes with a known defective rate of 2%. We want to find the probability that the batch will be accepted, meaning that there are less than two defective syringes in the sample of ten.
To solve this problem, we can use the binomial probability formula:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of having exactly x successes (in this case, the number of defective syringes in the sample)
n is the sample size (in this case, ten syringes)
x is the number of defective syringes found in the sample
p is the probability of success (in this case, the defective rate of 2% or 0.02)
C(n, x) is the binomial coefficient, which calculates the number of ways to choose x successes out of n trials. It can be calculated as C(n, x) = n! / (x! * (n-x)!)
In this case, we want to find the probability of having 0 or 1 defective syringes in the sample, which means:
P(x<2) = P(x=0) + P(x=1)
Let's calculate it step by step:
P(x=0) = C(10, 0) * 0.02^0 * (1-0.02)^(10-0) = 1 * 1 * 0.98^10 = 0.817
P(x=1) = C(10, 1) * 0.02^1 * (1-0.02)^(10-1) = 10 * 0.02 * 0.98^9 = 0.176
P(x<2) = P(x=0) + P(x=1) = 0.817 + 0.176 = 0.993 (rounded to three decimal places)
Therefore, the correct answer is 0.993, which is not among the answer choices provided.