the quality control inspector of a production plant will reflect a batch of syringes if two or more defectives syringes are found in a random sample of ten syringes taken from the batch of two percent defective syringes.what is the probability that the batch will be accepted .round your answer to three decimal places.
The easiest way to do these kinds of problems is to use a binomial probability table. If you use a table, you should have values for x, n, and p.
Use x = 0, 1
n = 10
p = .02 (for 2%)
Once you have the probabilities, add them together. Round your answer as requested.
To find the probability that the batch will be accepted, we need to calculate the probability of having two or fewer defective syringes in a random sample of ten syringes taken from the batch.
First, we need to find the probability of getting exactly two defective syringes in the sample.
The probability of getting a defective syringe in one trial is 2% or 0.02. Therefore, the probability of getting exactly two defective syringes in a random sample of ten syringes can be calculated using the binomial probability formula:
P(X=k) = (n C k) * p^k * (1-p)^(n-k)
Where:
P(X=k) represents the probability of getting exactly k defective syringes
(n C k) is the binomial coefficient, which represents the number of ways to choose k defective syringes from the sample of size n
p is the probability of getting a defective syringe in one trial, which is 0.02
k is the number of defective syringes
n is the sample size, which is 10
Using this formula, we can calculate the probability of getting exactly two defective syringes in the sample:
P(X=2) = (10 C 2) * (0.02)^2 * (1-0.02)^(10-2)
Calculating this expression:
P(X=2) = (10! / (2!(10-2)!)) * (0.02)^2 * (0.98)^8
P(X=2) = (45) * (0.02)^2 * (0.98)^8
P(X=2) ≈ 0.0432
Next, we need to find the probability of getting fewer than or equal to two defective syringes. This can be calculated by summing the probabilities of getting exactly 0, 1, and 2 defective syringes:
P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)
To calculate P(X=0) and P(X=1), we can use the same formula as before with k=0 and k=1, respectively.
P(X=0) = (10 C 0) * (0.02)^0 * (1-0.02)^(10-0)
P(X=0) = (1) * (0.02)^0 * (0.98)^10
P(X=0) ≈ 0.817
P(X=1) = (10 C 1) * (0.02)^1 * (1-0.02)^(10-1)
P(X=1) = (10) * (0.02)^1 * (0.98)^9
P(X=1) ≈ 0.302
P(X ≤ 2) = 0.817 + 0.302 + 0.0432
P(X ≤ 2) ≈ 0.1622
Therefore, the probability that the batch will be accepted is approximately 0.162 rounded to three decimal places.