A box contains 10 items, of which 3 are defective and 7 are non-defective. Two items are randomly selected, one at a time, with replacement, and x is the number of defectives in the sample of two. Explain why x is a binomial random variable.
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For the first question about the box containing 10 items of which 3 are defective, what is the probability of any one selected item being non-defective?
X
There are n specified ‘trials’.
Each observation results in one of two possible outcomes, called ‘success’ and ‘failure’.
The probability of a ‘success’ remains the same from one trial to the next.
The outcomes are independent from one trial to the next
7/10 = 0.7
x is a binomial random variable because it meets the following conditions that define a binomial random variable:
1. The experiment consists of a fixed number of identical trials, in this case, two trials of selecting items from the box.
2. Each trial has two possible outcomes, defective or non-defective.
3. The probability of success (selecting a defective item) remains the same for each trial.
4. The trials are independent, meaning that the outcome of one trial does not affect the outcome of another trial.
For the first question, the probability of any one selected item being non-defective can be calculated as follows:
Probability of selecting a non-defective item = (Number of non-defective items) / (Total number of items)
In this case, there are 7 non-defective items out of a total of 10 items.
Probability of selecting a non-defective item = 7/10 = 0.7 or 70%
To understand why x is a binomial random variable, let's first define what a binomial random variable is. A binomial random variable is a random variable that counts the number of successes in a fixed number of independent Bernoulli trials, where each trial has the same probability of success.
In this scenario, we are conducting two independent trials, as we are selecting two items from the box. Each item has the same probability of being defective or non-defective, regardless of the outcome of the previous trial. This is because we are sampling with replacement, which means that after each trial, the item is placed back into the box and the probabilities remain the same.
The number of defectives in the sample of two, denoted by x, can take on one of three values: 0, 1, or 2. The probability of having 0 defectives is calculated by multiplying the probability of selecting a non-defective item in both trials:
P(x = 0) = P(non-defective) * P(non-defective) = (7/10) * (7/10)
Similarly, the probability of having 1 defective is calculated by multiplying the probability of selecting one non-defective and one defective item, in any order:
P(x = 1) = [P(non-defective) * P(defective)] + [P(defective) * P(non-defective)]
= [(7/10) * (3/10)] + [(3/10) * (7/10)]
Finally, the probability of having 2 defectives is calculated by multiplying the probability of selecting a defective item in both trials:
P(x = 2) = P(defective) * P(defective) = (3/10) * (3/10)
Since the probability of each outcome can be calculated using a fixed number of independent trials with the same probability of success, x is a binomial random variable.
Now, let's move on to the second question about the probability of any one selected item being non-defective. Since there are 10 items in total and 7 of them are non-defective, we can find the probability by dividing the number of non-defective items by the total number of items:
Probability of a selected item being non-defective = Number of non-defective items / Total number of items
= 7 / 10
= 0.7