# Thread: Probability of biochemical experiment, distribution

1. ## Probability of biochemical experiment, distribution

In a biochemical experiment, n organisms are placed in a nutrient medium, and the number of organisms X which survive for a given period is recorded. The probability distribution of X is assumed to be given by

$\displaystyle P(X=k)=\frac{2(k+1)}{(n+1)(n+2)}\ for\ 0 \leq k \leq n,$

and $\displaystyle 0$ otherwise

1) Calculate the probability that at most a proportion $\displaystyle \alpha$ of the organisms survive, and deduce that for large n this is approximately $\displaystyle \alpha^2$

2) Find the smallest value of $\displaystyle n$ for which the probability of there being at least one survivor among the n organisms is at least 0.95

2. Originally Posted by acevipa
In a biochemical experiment, n organisms are placed in a nutrient medium, and the number of organisms X which survive for a given period is recorded. The probability distribution of X is assumed to be given by

$\displaystyle P(X=k)=\frac{2(k+1)}{(n+1)(n+2)}\ for\ 0 \leq k \leq n,$

and $\displaystyle 0$ otherwise

1) Calculate the probability that at most a proportion $\displaystyle \alpha$ of the organisms survive, and deduce that for large n this is approximately $\displaystyle \alpha^2$

2) Find the smallest value of $\displaystyle n$ for which the probability of there being at least one survivor among the n organisms is at least 0.95
1) Probability that at most a proportion $\displaystyle \alpha$ survives is the same as the probability than at most the number that survive is $\displaystyle \lfloor \alpha n \rfloor$ survive

This is:

$\displaystyle p=\sum_{k=0}^{\lfloor \alpha n \rfloor} P(X=k)$

and the rest is just algebra.

CB