两枚偏骰的三种抽样方案：方差比较

The standard binomial sampling scheme assumes n independent trials with a constant probability of success. Suppose, for example, that we are given a single fair die, and that the success event consists in throwing, say, a 5. Thus, each single throw results in a success with probability $\mathrm{p} = \frac{1}{6}$ . Within $\mathrm{n} = 144$ independent trials, we would thus expect about $\mathrm{n} \cdot \mathrm{p} = 24$ successes, and the variance of the number of successes would be equal to $\mathrm{n} \cdot \mathrm{p} \cdot (1 - \mathrm{p}) = 20$ .

a. Suppose we are given two biased dice. Die A will show a 5 with probability $\mathrm{p} = \frac{1}{4}$ , and die B shows a 5 with probability $\mathrm{p} = \frac{1}{12}$ . With these two dice 144 trials are conducted, as follows. For the first 72 trials we use dieA, and during the last 72 trials we use die B. Given that the average success probability across both dice is equal to $\left(\frac{1}{4} + \frac{1}{12}\right) / 2 = \frac{1}{6}$ , we would again expect 24 successes. Is the variance of the number of successes larger than, equal to, or smaller than with the standard binomial sampling scheme?

b. Again we are given the two dice A and B described in a. This time, however, we select at random one die and then throw this selected die 144 times, observing again the number of throws yielding a 5. Is the variance of the number of successes larger than, equal to, or smaller than with the standard binomial sampling scheme?

c. Once again we are given the two dice A and B. This time, however, in each of 144 trials we start by choosing at random one of the two dice, then throw it and observe whether or not the trial yields a 5. Is the variance of the number of successes larger than, equal to, or smaller than with the standard binomial sampling scheme?