例 2.2.4 (三胞胎出生概率)

假如一位孕妇生三胞胎的概率为 ${10}^{-4}$, 求在 100000 个孕妇中有 0, 1, 2 次生三胞胎的概率.

解

按二项分布的 $n=100000$ 和 $p={10}^{-4}$, 并用 $\mathrm{R}$ 软件计算有 $b\left(0,100000,0.0001\right)=\mathrm{dbinom}\left(0,100000,0.0001\right)=4.537723\times {10}^{-5}$ b\left( {1, {100000}, {0.0001}}\right) = \operatorname{d$binom}\left( {1, {100000}, {0.0001}}\right) = {0.0004538177} $b\left(2,100000,0.0001\right)=\mathrm{dbinom}\left(2,100000,0.0001\right)=0.002269293$

再按泊松逼近, $\lambda =np=10$ 和 $\mathrm{R}$ 软件计算有

$$b\left(0,100000,0.0001\right)=\text{ dpois }\left(0,10\right)=4.539993\times {10}^{-5},$$ $$b\left(1,100000,0.0001\right)=\text{ dpois }\left(1,10\right)=0.0004539993,$$ $$b\left(2,100000,0.0001\right)=\mathrm{dpois}\left(2,10\right)=0.002269996.$$

可见, 这里用泊松分布逼近二项分布的近似程度很令人满意.