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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.

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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-

This problem has been solved

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