Bobo the amoeba currently lives alone in a pond After one mi

Bobo, the amoeba, currently lives alone in a pond After one minute Bobo will either die, split into two amoebas, or stay the same, with equal probability. Find the variance for the number of amoebas in the bond after one minute.

Solution

let X be the random variable denoting the number of amoebas in the pond after one minute.

if bobo dies then X=0

if bobo stays the same then X=1

if bobo splits into two then X=2

each have the same probability

so probability distribution of X is

X:             0                        1                                2

P[X=x]:           1/3                      1/3                            1/3

so V[X]=E[X²]-E[X]²

now E[X]=0*1/3+1*1/3+2*1/3=1

E[X²]=0*1/3+1*1/3+4*1/3=5/3

so variance of X is V[X]=5/3-1²=2/3 [answer]