ProbabilityHardBloom L3

Question

A certain kind of bacteria either die, split into two or split into three bacteria. All splits are exact copies. The probability of dying is $\frac{1}{4}$, the probability of splitting into two is $\frac{1}{2}$ and splitting into three is $\frac{1}{4}$.

The probability that it survives for infinite length of time is

Options

A.$\frac{5 - \sqrt{13}}{2}$
B.$\frac{4}{7}$
C.$\frac{\sqrt{13} - 3}{2}$
D.$\frac{6 - \sqrt{13}}{2}$

Solution

Sol. [Key Concept: Bacteria die/split: solve the fixed-point equation for survival.]

Let probability of single bacteria to die $= P$

$$\therefore\ P = \frac{1}{4} \times 1 + \frac{1}{2} \times P \times P + \frac{1}{4} \times P \times P \times P$$

bacteria die or split

A → bacteria do not split

B → bacteria split into 2

$C \rightarrow$ bacteria split into 3

D → bacterial die

$$\begin{matrix} & & \ \ \ \ \ \ \ \ & P(D) & \ = P(D \cap \text{ }A) + P(D \cap \text{ }B) + P(D \cap C) \\ & & \ \ \ \ \ \ \ \ & P & \ = \frac{1}{4} \times 1 + \frac{1}{2} \times P \times P + \frac{1}{4} \times P \times P \times P \\ & \ \Rightarrow & \ \ \ \ \ \ \ \ & P^{3} + 2P^{2} - 4P + 1 & \ = 0 \\ & \ \Rightarrow & \ \ \ \ \ \ \ \ & \ (P - 1)\left( P^{2} + 3P - 1 \right) & \ = 0 \\ & \ \Rightarrow & \ \ \ \ \ \ \ \ & P & \ = \frac{- 3 + \sqrt{13}}{2} \end{matrix}$$

∴ Probability that bacteria survive

$$= 1 - \frac{\sqrt{13} - 3}{2} = \frac{5 - \sqrt{13}}{2}$$

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