suppose a manufacterer of transistors believes that, on the average, one defective transistor occurs in every 100 transistors.
a. find the probability that a batch of 20 transistors has 2 defectives.
b. find the probability that a batch of 20 transistors has at most 2 defectives.
You can use something called the binomial probability function for these kinds of problems.
The function is this:
P(x) = (nCx)(p^x)[q^(n-x)]
For a:
n = 20
p = 1/100 = .01
q = 1 - p = 1 - .01 = .99
Substituting:
P(2) = (20C2)(.01^2)(.99^18) = (190)(.0001)(.8345) = .0159
For b:
n = 20
p = .01
q = .99
x = 0, 1, 2
Find P(0), P(1), and P(2). Add all values together for your probability.
I hope this helps.
To find the probability that a batch of 20 transistors has 2 defectives, you can use the binomial probability function.
The function is: P(x) = (nCx)(p^x)(q^(n-x)), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure.
For this problem:
n = 20 (since there are 20 transistors in the batch)
p = 1/100 = 0.01 (since the probability of a defective transistor is 1/100)
q = 1 - p = 1 - 0.01 = 0.99 (since q represents the probability of a non-defective transistor)
Substituting these values into the formula:
P(2) = (20C2)(0.01^2)(0.99^(20-2))
Now, let's calculate each component:
(20C2) = 20! / (2!(20-2)!) = 20! / (2! * 18!) = (20 * 19) / 2 = 190
(0.01^2) = 0.0001
(0.99^(20-2)) = 0.99^18 = 0.8345
Now, we can substitute these values back into the formula:
P(2) = (190)(0.0001)(0.8345) = 0.0159
Therefore, the probability that a batch of 20 transistors has 2 defectives is 0.0159.
To find the probability that a batch of 20 transistors has at most 2 defectives, you need to find the total probability for having 0, 1, and 2 defectives.
Let's calculate each component:
P(0) = (20C0)(0.01^0)(0.99^(20-0)) = 1(1)(0.99^20) = 0.8171
P(1) = (20C1)(0.01^1)(0.99^(20-1)) = 20(0.01)(0.99^19) = 0.1809
P(2) = (20C2)(0.01^2)(0.99^(20-2)) = 190(0.0001)(0.99^18) = 0.0159
Adding all these probabilities together:
P(at most 2) = P(0) + P(1) + P(2) = 0.8171 + 0.1809 + 0.0159 = 1
Therefore, the probability that a batch of 20 transistors has at most 2 defectives is 1.