关于能否使用binomial distribution求解次品微机采购概率分布的疑问
Great question! Let’s break this down clearly to understand why binomial distribution isn’t the right fit here, and what we should use instead.
First, let’s recap the core requirements for binomial distribution to apply:
- Independent trials: The outcome of one pick doesn’t affect the next.
- Constant probability: The chance of picking a defective computer stays the same for every trial.
- Fixed number of trials (in this case, 2 purchases).
In your problem, we’re dealing with a small, finite population (8 total computers) and we’re making without-replacement picks—once you buy a computer, it’s removed from the pool. This breaks both key binomial assumptions:
- If you pick a non-defective first, the probability of picking a defective second becomes 3/7 (not the original 3/8).
- If you pick a defective first, the probability of picking another defective drops to 2/7.
Since the probability changes with each pick, binomial distribution isn’t appropriate here. Instead, we use the hypergeometric distribution for without-replacement sampling from a finite population.
Let’s calculate the probability distribution for (x) (number of defectives) using the hypergeometric formula:
[
P(X=k) = \frac{\binom{K}{k} \times \binom{N-K}{n-k}}{\binom{N}{n}}
]
Where:
- (N = 8) (total computers), (K = 3) (defective computers), (n = 2) (number purchased), (k) = number of defectives.
Calculating each case:
- (x=0) (no defectives):
[
P(X=0) = \frac{\binom{3}{0} \times \binom{5}{2}}{\binom{8}{2}} = \frac{1 \times 10}{28} = \frac{5}{14} \approx 0.357
] - (x=1) (one defective):
[
P(X=1) = \frac{\binom{3}{1} \times \binom{5}{1}}{\binom{8}{2}} = \frac{3 \times 5}{28} = \frac{15}{28} \approx 0.536
] - (x=2) (two defectives):
[
P(X=2) = \frac{\binom{3}{2} \times \binom{5}{0}}{\binom{8}{2}} = \frac{3 \times 1}{28} = \frac{3}{28} \approx 0.107
]
A quick side note: If the population were extremely large (say, 10,000 computers instead of 8), the change in probability after each pick would be negligible, and binomial distribution could be used as a rough approximation. But for small populations like this, hypergeometric gives the exact probabilities.
备注：内容来源于stack exchange，提问作者Numa ARX
