你能尝出 M&M 的颜色吗

把“随机猜测”作为原假设 $H_0$：每次答对的概率 $p=1/5=0.2$，且各次独立，则正确数 $X\sim\mathrm{Bin}(n,0.2)$。

实验 1（n=3,k=2）：计算在随机猜测下至少答对 2 次的概率（p 值）

$$\mathbb{P}(X\ge 2)=\binom{3}{2}(0.2)^2(0.8)+(0.2)^3=0.104.$$

该概率并不小，因此样本太少，无法据此认为他真的能凭味道辨色。

实验 2（n=100,k=40）：在 $H_0$ 下期望为 $\mathbb{E}[X]=20$，标准差为

$$\sqrt{np(1-p)}=\sqrt{100\cdot 0.2\cdot 0.8}=4.$$

观测到 40 比期望高出 20，相当于约 5 个标准差，p 值极小（远小于 0.001）。

因此第二次实验对“随机猜测”提供了很强的反证：从统计角度应认为其辨色能力显著高于瞎猜（当然也需要排除作弊或实验设计漏洞）。

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英文解析

Take "random guessing" as the null hypothesis $H_0$: the probability of being correct each time is $p=1/5=0.2$, and trials are independent, so the number of correct answers $X\sim\mathrm{Bin}(n,0.2)$.

Experiment 1 ($n=3, k=2$): Calculate the probability (p-value) of getting at least 2 correct under random guessing.

$$\mathbb{P}(X\ge 2)=\binom{3}{2}(0.2)^2(0.8)+(0.2)^3=0.104.$$

This probability is not small; therefore, the sample size is too small to conclude that he can indeed distinguish colors by taste.

Experiment 2 ($n=100, k=40$): Under$H_0$, the expected value is$\mathbb{E}[X]=20$, and the standard deviation is

$$\sqrt{np(1-p)}=\sqrt{100\cdot 0.2\cdot 0.8}=4.$$

Observing 40 is 20 units higher than the expectation, which corresponds to approximately 5 standard deviations. The p-value is extremely small (much less than 0.001).

Thus, the second experiment provides strong evidence against "random guessing": statistically, one should conclude that their ability to distinguish colors is significantly better than guessing (of course, one must also rule out cheating or experimental design flaws).