A computer randomly generates two thousand numbers from this list: 1, 3, 5, 7, 9, and 11. Predict how many of the randomly gener
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Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.
In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.
<h3>What is the hypergeometric distribution formula?</h3>The formula is:
The parameters are:
- x is the number of successes.
- N is the size of the population.
- n is the size of the sample.
- k is the total number of desired outcomes.
In this problem:
- There are 12 bulbs, hence N = 12.
- 3 are defective, hence k = 3.
The third defective bulb is the fifth bulb if:
- Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
- The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.
Hence:
0.2182 x 1/8 = 0.0273.
0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.
More can be learned about the hypergeometric distribution at brainly.com/question/24826394
Answer:
105 minutes
Step-by-step explanation:
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