# Simple probability problem

1. May 9, 2012

### chuy52506

1. The problem statement, all variables and given/known data

Suppose i have 150 birds each having a survival rate of .5.

2. Relevant equations

What would be the probability that there are 0 birds after 5 times steps?

3. The attempt at a solution

The only way I could think of doing this would be to find the expected value after four time steps which would be 150*(.5)^4 = 4.6875. This would indicate that there are 4 birds left so I would find the probability 4 birds die in one time step which is .5^4. However, it is incorrect..

Last edited: May 9, 2012
2. May 9, 2012

### chuy52506

Anyone have an idea on how to start this?

3. May 9, 2012

### Ray Vickson

If by survival rate you mean probability of survival, then you could model the problem as a Markov chain on the state space {0,1,2,...,150}. At any integer time point the current state is the number of birds still alive at that time. We start at time t = 0 in state 150. Whenever the state is i (i >= 1) the number of birds that will die during the next time step is a binomial random variable with parameters i and 0.5, so P{k birds die by time t+1|i alive at time t} = C(i,k)/2^i, for k = 0,1,2, ..., i, where C(a,b) is the binomial coefficient C(a,b) = a!/[b! (a-b)!]. If k birds die there will be i-k left alive at time t+1.

You can work out the 150x150 one-step transition probability matrix; its entries are binomial probabilities. You want the probability of being in state 0 at time 5. Since state 0 is "absorbing" (meaning that once you reach 0 you stay in state 0 forever), that means that you reach state 0 for the first time at t <= 5. The answer will just be the (0,150) element of the matrix P^5, the fifth power of P in the sense of matrix multiplication.

I would not call this a simple problem.

RGV

4. May 10, 2012

### Ray Vickson

Instead of the method in my previous response, it is easier to tackle the problem directly, and using a generating function approach gives a surprisingly simple answer. Start with N birds. The number who die in the first time step is X1 = binomial(N,1/2), whose generating function is f1(z) = sum_{k=0..N} z^k*C(N,k)/2^N = (1/2 + 1/2 * z)^N. The number who die by the end of the second period is X1 + X2, where X1 = binomial(N,1/2), and X2 (given X1) is binomial(N-X1,1/2). The generating function of X1+X2 is f2(z) = E[z^(X1+X2)] = sum_{k} P{X1=k} E[z^(X1+X2)|X1=k] = sum_k P{X1=k} z^k * (1/2 + 1/2 * z)^(N-k) = ((1+z)/2)^N * f1(2*z/(1+z)) = [(1+z)^N /2^N](1/2 + 1/2 * 2*z/(1+z))^N = (1/4 + 3/4 * z)^N.

Note: this means that the P{X1+X2=k} = C(N,k)(3/4)^k (1/4)^(N-k).

In a similar way we can get the generating function fn(z) of X1+X2+...+Xn, and it is much simpler than I would ever have imagined. The final answer is simple, but I still would not call this a simple problem.

Note added in editing: to contradict the statement I made above---the problem IS simple if you look at it the right way. The trick is to look at the birds that survive, not the ones that die.

RGV

Last edited: May 10, 2012
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