1. Consider any one lineage from a single starting cell given the context of the question, then using total probability:
Let \( p_0 \) = P(0 offspring)
Let \( p_1 \) = P(1 offspring)
Let \( p_2 \) = P(2 offspring)
P(extinction) = P(extinction | starting cell produced 0 offspring) * \( p_0 \) + P(extinction | starting cell produced 1 offspring) * \( p_1 \) + P(extinction | starting cell produced 2 offspring) * \( p_2 \)
Let \( q \) = P(extinction) then P(extinction | starting cell produced 1 offspring) = \( q \) and P(extinction | starting cell produced 2 offspring) = \( q^2 \).
Thus \( q \) = \( p_0 \) + \( q \) * \( p_1 \) + \( q^2 \) * \( p_2 \). In this case \( q = 0.2 + 0.2q + 0.6q^2 \) which is a quadratic that can be solved easily. The two roots are 1 and \( 1/3 \), the latter being the probability in question. Note that there will always be a root with value 1, if there is also a root less than 1 in magnitude then this is the probability, otherwise the ultimate extinction probability is 1. We could also tell that the extinction probability must be strictly less than 1 as the expected number of children per cell is greater than 1. Given that there are two starting cells, the overall probability in question is thus \( q^2 = 1/9 \).
2. P(gen_3 = 0 | gen_2 > 0) = P(gen_3 = 0 and gen_2 >0) / P(gen_2 > 0).
Method 1 - understanding all the possibilities
The numerator can be found by summing the probabilities for all paths that meet the criteria so:
Gen_ gen_0 -> gen_1 -> gen_2 -> gen_3
Path 1. 1--------->1--------->1--------->0
Path 2. 1--------->1--------->2--------->0
Path 3. 1--------->2--------->1 (1,0)-->0
Path 4. 1--------->2--------->2 (1,1)-->0
Path 5. 1--------->2--------->2 (2,0)-->0
Path 6. 1--------->2--------->3 (2,1)-->0
Path 7. 1--------->2--------->4 (2,2)-->0
Where, for clarity, I have indicated in gen_2 the offspring combinations for the total number in that generation which will help us get the probabilities correct. We should also be careful with paths 3, 5 and 6 which have 2 ways for them to occur. For example, in path 3 that goes from 2 cells to 1 cell between gen_1 and gen_2, it could be either of the 2 cells that produced that one offspring, the other producing none.
\( p(1) = 0.2^3 = 1/125 \).
\( p(2) = 0.2 * 0.6 * 0.2^2 = 3/625 \).
\( p(3) = 0.6 * 0.2^2 * 2 * 0.2 = 6/625 \).
....
Calculating and summing all of these you should end up with a total probability for the numerator of \( 2392/78125. \)
The denominator, \( p(gen_2 \neq 0) = 0.2*0.8 + 0.6* (1-(0.2^2)) = 92/125 \). This is from P(gen_2 \( \neq \) 0) = P(gen_2 \( \neq \) 0 | gen_1 = 1) * P(gen_1 = 1) + P(gen_2 \( \neq \) 0 | gen_1 = 2) * P(gen_1 = 2). Ignoring the case when gen_1 = 0 as this has 0 probability of gen_2 then being not extinct.
Overall the desired probability is \( (2392/78125) / (92/125) = 26/625. \)
Method 2 - using extinction probabilities
Similar to the equation we used in question 1, if we let \( q_m \) = P(gen_m = 0) then we can see that:
\( q_m = p_0 + p_1 * q_{m-1} + p_2 * (q_{m-1})^2 \).
This is from the idea that if a population is to be extinct in m generations, then the offspring must have been extinct in (m-1) generations (note this not the probability of going extinct in exactly gen_m but by gen_m). To answer the question we need \( q_0, q_1, q_2 \) and \( q_3 \).
\( q_0 = 0 \)
\( q_1 = p_0 = 0.2 \)
\( q_2 = p_0 * (1 + p_1 + p_2*p_0) =33/125 \)
\( q_3 = 0.2946176 \)
The probability in question is \( (q_3 - q_2 ) / (1 - q_2) = 26/625 \).
