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there is an amoeba....in every minute it has four options : (0) it dies (1) it remains as it is (2) it doubles (3) it triples
find the probability that after 2 minutes the amoeba dies ....(.i.e. if it doubles or triples all of them will die individually)....all the chances are equiprobable to the amoeba !
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7 Answers
62Probability of dying in 1 minute = 1/4
Probability of dying in 2nd minute = 1/4x1/4 + 1/4 x(1/16) + 1/4(1/64) = 1/16 ( 1+ 1/4 + 1/16) = 21/256
Prob of dying in 2 minutes = 85/256
Prob of dying after 2 minutes = 1-85/256 = 171/256
I Hope i have not made any calculation mistakes :)
3sir i have a dbt
isnt the probability of the amoeba to die in the 2nd minute=3/4.1/4
so the reqd probability is 1-3/16
62that is what i also did the first time
read teh question again
the question says that the amoeba gets divided in 2 parts
If one of them dies.. then does the amoeba die?
62also even then the expression you have given is not correct
prob to die in 1st minute is 1/4
what you have done is prob of not dying in the 2nd minute.
You have to calculate the prob of dying after then 2nd minute.. so you would have to subtract a further 1/4th from the answer you got.
3yeah u were rite i havent read the text inside the brackett.thanx for pointing out the mistake sir.
1same question but extended.
Q A jar begins with one amoeba. Every minute, every amoeba
turns into 0, 1, 2, or 3 amoebae with probability 25%
for each case ( dies, does nothing, splits into 2, or splits
into 3). What is the probability that the amoeba population
eventually dies out?
Sol:If p is the probability that a single amoeba's descendants will die
out eventually, the probability that N amoebas' descendents will all
die out eventually must be p^N, since each amoeba is independent of
every other amoeba. Also, the probability that a single amoeba's
descendants will die out must be independent of time when averaged
over all the possibilities. At t=0, the probability is p, at t=1 the
probability is 0.25(p^0+p^1+p^2+p^3), and these probabilities must be
equal. Extinction probability p is a root of f(p)=p. In this case,
p = sqrt(2)-1.
The generating function for the sequence P(n,i), which gives the
probability of i amoebas after n minutes, is f^n(x), where f^n(x) ==
f^(n-1) ( f(x) ), f^0(x) == x . That is, f^n is the nth composition
of f with itself.
Then f^n(0) gives the probability of 0 amoebas after n minutes, since
f^n(0) = P(n,0). We then note that:
f^(n+1)(x) = ( 1 + f^n(x) + (f^n(x))^2 + (f^n(x))^3 )/4
so that if f^(n+1)(0) -> f^n(0) we can solve the equation.
The generating function also gives an expression for the expectation
value of the number of amoebas after n minutes. This is d/dx(f^n(x))
evaluated at x=1. Using the chain rule we get f'(f^(n-1)(x))*d/dx(f^(n-1)(x))
and since f'(1) = 1.5 and f(1) = 1, we see that the result is just
1.5^n, as might be expected.
