Question

a lot of manufactured items contains 20 percent defectives. A sample of 10 items from this lot is chosen at random. what is the probability that the sample conatins at most 3 defective items ?

Answer 1

The probability that a sample of 10 items with 20 percent defectives will contain at most 3 defective items is calculated using the binomial distribution, summing up the probabilities of having 0, 1, 2, or 3 defects.

Step-by-step explanation:

The question is to find the probability that a sample of 10 items, which contains 20 percent defectives, will have at most 3 defective items. This is a typical problem that can be solved using the binomial distribution because the items can be categorized into two groups: defective and non-defective. For our binomial distribution, the probability of success (finding a defective item) is 0.2, and we have 10 trials (items). We want to find the probability of having 0, 1, 2, or 3 successes (defective items).

To compute this, we use the binomial probability formula:

`P(X = k) = C(n, k) . (p)^k . (1-p)^(n-k)`

where P(X = k) is the probability of k successes, C(n, k) is the combination of n items taken k at a time, p is the probability of a single success, and n is the number of trials.

Therefore, the total probability would be P(X ≤ 3) which is the sum of the probabilities of having 0, 1, 2, or 3 defective items out of the 10 sampled (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)).

Answer 2

To find the probability that a sample of 10 items contains at most 3 defective items, we can use the binomial distribution. The formula for this is P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where n is the sample size, p is the probability of a defective item, and k is the number of defective items. Calculate the probabilities for each value of k (0 to 3) and add them together to get the final answer.

Step-by-step explanation:

To solve this problem, we can use the binomial distribution.

The formula for the probability of getting k successes in n trials is P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where C(n, k) is the number of combinations of n objects taken k at a time and p is the probability of success.

In this case, n = 10 (the size of the sample) and p = 0.2 (the probability of a defective item).

We want to find the probability that the sample contains at most 3 defective items, so we need to find P(X=0) + P(X=1) + P(X=2) + P(X=3).

Using the formula, we can calculate:

P(X=0) = C(10, 0) * (0.2)^0 * (0.8)^10

P(X=1) = C(10, 1) * (0.2)^1 * (0.8)^9

P(X=2) = C(10, 2) * (0.2)^2 * (0.8)^8

P(X=3) = C(10, 3) * (0.2)^3 * (0.8)^7

Add these probabilities together to get the final answer.