Question

In Kerrville has 50 people left and they each have an 6% chance of being eaten each day, then what is the probability that more than 30 people survive for the 7 days required for the national guard to arrive on the scene?

Answer 1

It's a binomial distribution problem:

Probability of success(being survived) on a given day = p= 94%=0.94

Probability of failure (being eaten) =q =1- p = 6% =0.06

n = 50 people;

Required to compute [P(X>30)]⁷ , that is the probability of more than 30 people being survived for 7 days.

P(X>30) = 0.999999 for one day.

For 7 days, (0.999999)⁷ = 0.999993

Thus, the probability that more than 30 people survive for the 7 days required for the national guard to arrive on the scene is 0.999993