Is there anyone that can solve the below questions? This is a new topic for me A cell produces 0, 1 or 2 offspring with probabilities 0.2, 0.2 and 0.6 respectively. (You may assume that cells produce offspring independently of one another.) What is the extinction probability for a colony that initially consists of two cells? If a colony starts with a single cell, what is the probability that it is extinct in the third generation given that it did not die out in the second generation?
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 \( (q_3 - q_2 ) / (1 - q_2) = 26/625 \).
