A company manufactures caclculators in batchers of 64 and there is a 5% rate of defects. Find the probability of getting exactly three defects in a batch
I think I remember a bit of this from when we were generating OC curves for a math modeling project last semester.
In my notes there is the equation:
P(d)= (n!/(d!(n-d)!))*(p^d)(1-p)^(n-d)
It's assumed that the total lot is very large, so when you take n, the sample size, out doesn't change the remainder.
P(d)= prop of observing exactly d
number of defectives= d
random sample= n
p= fraction of defectives per lot.
Hope this helps
To find the probability of getting exactly three defects in a batch, we can use the binomial probability formula. The formula for the probability of x successes in n independent trials, where the probability of success on each trial is p, is:
P(x) = (nCx) * p^x * (1-p)^(n-x)
In this case, n (the number of trials) is 64 (the batch size), p (the probability of success on each trial) is 0.05 (the defect rate), and x (the number of successes) is 3 (the number of defects we want to find the probability for).
Now let's plug in the values and calculate the probability:
P(3) = (64C3) * 0.05^3 * (1-0.05)^(64-3)
First, let's calculate the binomial coefficient (nCx), which represents the number of ways to choose x items from a set of n items. The formula for the binomial coefficient is:
(nCx) = n! / (x! * (n-x)!)
Using this formula, we can calculate:
(64C3) = 64! / (3! * (64-3)!)
Simplifying further:
(64C3) = 64! / (3! * 61!)
Next, let's calculate the probability:
P(3) = (64C3) * 0.05^3 * (1-0.05)^(64-3)
Calculating (64C3):
(64C3) = 64! / (3! * 61!)
(64C3) = (64 * 63 * 62) / (3 * 2 * 1)
(64C3) = 416,128
Plugging in the values:
P(3) = 416,128 * 0.05^3 * (1-0.05)^(64-3)
P(3) = 416,128 * 0.05^3 * (0.95)^(61)
P(3) = 416,128 * 0.05^3 * 0.095
P(3) ≈ 0.0987
Therefore, the probability of getting exactly three defects in a batch is approximately 0.0987, or 9.87%.