A case of wine has 10 bottles, 2 of which contain spoiled wine. A sample of 3 bottles is
randomly selected from the case.
(a)What is probability that the sample contains 2 bottles of spoiled wine?
(b)What is probability that all 3 of the sampled bottles are spoiled?
(c)What are the mean and variance of the number of spoiled bottles in the sample?
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I don't know can you tell me please
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To answer these questions, we first need to understand the concept of probability and how to calculate it in this scenario.
Probability is a measure of the likelihood that a specific event will occur. In this case, we want to calculate the probability of different outcomes when selecting a sample of 3 bottles from a case of 10, where 2 bottles are spoiled.
(a) Probability of selecting 2 spoiled bottles:
To calculate this probability, we need to consider the number of ways we can select 2 spoiled bottles out of the 10 bottles in the case.
The number of ways to choose 2 spoiled bottles out of the 2 spoiled bottles in the case is 2C2, which is equal to 1.
The number of ways to choose the remaining 1 non-spoiled bottle out of the 8 non-spoiled bottles in the case is 8C1, which is equal to 8.
The total number of ways to choose any 3 bottles from the case is 10C3, which is equal to 120.
Therefore, the probability of selecting 2 bottles of spoiled wine from the sample is:
P(2 spoiled bottles) = (2C2 * 8C1) / 10C3 = (1 * 8) / 120 = 8/120 = 1/15 ≈ 0.067.
So, the probability that the sample contains 2 bottles of spoiled wine is approximately 0.067.
(b) Probability that all 3 sampled bottles are spoiled:
To calculate this probability, we need to consider the number of ways we can select all 3 spoiled bottles out of the 2 spoiled bottles in the case.
The number of ways to choose all 3 spoiled bottles out of the 2 spoiled bottles in the case is 2C3, which is equal to 0 since we cannot choose more than what is available.
Therefore, the probability of selecting all 3 spoiled bottles from the sample is:
P(all 3 spoiled bottles) = 0.
So, the probability that all 3 of the sampled bottles are spoiled is 0.
(c) Mean and variance of the number of spoiled bottles in the sample:
To calculate the mean and variance, we need to determine the probability distribution of the number of spoiled bottles in the sample.
Let's consider all possible outcomes when selecting a sample of 3 bottles:
- 0 spoiled bottles: This can only happen if all 3 bottles in the sample are non-spoiled. The probability of this is (8C3)/(10C3) = 56/120 = 7/15 ≈ 0.467.
- 1 spoiled bottle: This can happen if 2 out of the 3 bottles in the sample are non-spoiled and 1 is spoiled. The probability of this is (2C1 * 8C2)/(10C3) = (2 * 28)/120 = 56/120 = 7/15 ≈ 0.467.
- 2 spoiled bottles: This can happen if 1 out of the 3 bottles in the sample is non-spoiled and 2 are spoiled. The probability of this is (2C2 * 8C1)/(10C3) = (1 * 8)/120 = 8/120 = 1/15 ≈ 0.067.
The mean (expected value) is calculated as the sum of each possible outcome multiplied by its respective probability. Therefore, the mean of the number of spoiled bottles in the sample is:
Mean = (0 * 0.467) + (1 * 0.467) + (2 * 0.067) = 0 + 0.467 + 0.134 = 0.601.
The variance is calculated by the sum of each possible outcome's squared deviation from the mean, multiplied by its respective probability. Therefore, the variance of the number of spoiled bottles in the sample is:
Variance = (0 - 0.601)^2 * 0.467 + (1 - 0.601)^2 * 0.467 + (2 - 0.601)^2 * 0.067 ≈ 0.481.
So, the mean of the number of spoiled bottles in the sample is approximately 0.601, and the variance is approximately 0.481.