细菌存活 II

设灭绝概率为 $x$。对初始细菌的第一次结果分情况：

$$x=\frac14\cdot 1+\frac14 x+\frac14 x^2+\frac14 x^3.$$

化简得到

$$x^3+x^2-3x+1=0=(x-1)(x^2+2x-1).$$

在 $(0,1)$ 内的根为 $x=\sqrt{2}-1$。

因此 $a=2,b=1$，$a+b=3$。

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Original Explanation

Let $x$ be the probability that the bacterial population will die out. There are 4 total possibilities for the first bacterium: either it dies, stays the same, splits into two, or splits into three. In the first case, the bacterial population dies with probability 1. In the second case, the bacterial population dies with probability $x$, since the bacterial population dies when the bacterium that stays the same dies, which happens with probability $x$. In the third case, the bacterial population dies with probability $x^2$, as both bacteria that are formed must die in order for the population to die. In the fourth and final case, the bacterial population dies with probability $x^3$, as all three bacteria that are formed must die in order for the population to die. Thus, we can write $$x = \frac{1}{4} \times 1 + \frac{1}{4}x + \frac{1}{4}x^2 + \frac{1}{4}x^3 = \sqrt{2} - 1 \approx 0.41$$ Therefore, our answer is $2 + 1 = 3$.