There is a container having same type of small products. There is a task to select one component at a time. The selected component me be either defective or non-defective. I The probability for a product to be non-defective is 0.9 then determine the probability that: a. The 4th non defective product is found at 9th trial. b. The 1st non defective product is found at 4th trial.
Question
There is a container having same type of small products. There is a task to select one component at a time. The selected component me be either defective or non-defective. I The probability for a product to be non-defective is 0.9 then determine the probability that: a. The 4th non defective product is found at 9th trial. b. The 1st non defective product is found at 4th trial.
Solution
The problem can be solved using the Negative Binomial Distribution. The Negative Binomial Distribution gives the probability of k-1 successes and 1 failure in n trials. The failure must occur on the nth trial.
a. The 4th non-defective product is found at the 9th trial.
This means that we have 3 non-defective products in the first 8 trials and the 4th non-defective product in the 9th trial. The probability of a product being non-defective is 0.9 and defective is 0.1.
Using the formula for Negative Binomial Distribution:
P(X = n) = C(n-1, k-1) * (p^k) * ((1-p)^(n-k))
where: n = number of trials = 9 k = number of successes = 4 p = probability of success = 0.9
P(X = 9) = C(9-1, 4-
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