Suppose there are 300,000,000 people each at an independent risk of 1/10,000,000 of dying. What is the probability that exactly 30 people die (which is the expected value)?

Guest Aug 18, 2015

#1**+10 **

Probability of an individual dying: p = 10^{-7}

Probability of an individual surviving: q = 1-p

Number of individuals: N = 3*10^{8}

Probability of exactly 30 individuals dying = (probability of 30 dying)*(probability of N-30 surviving)*(number of ways of choosing 30 out of N) = p^{30}*q^{N-30}*ncr(N,30) ≈ 0.073

(This should be confirmed, or otherwise, by someone who actually knows something about probability!)

.

Alan Aug 19, 2015

#1**+10 **

Best Answer

Probability of an individual dying: p = 10^{-7}

Probability of an individual surviving: q = 1-p

Number of individuals: N = 3*10^{8}

Probability of exactly 30 individuals dying = (probability of 30 dying)*(probability of N-30 surviving)*(number of ways of choosing 30 out of N) = p^{30}*q^{N-30}*ncr(N,30) ≈ 0.073

(This should be confirmed, or otherwise, by someone who actually knows something about probability!)

.

Alan Aug 19, 2015

#2**+5 **

I assume that this is for some given time period ?

Suppose there are 300,000,000 people each at an independent risk of 1/10,000,000 of dying. What is the probability that exactly 30 people die (which is the expected value)?

30 people die 299999970 live

Prob of dying = 10^(-7)

prob of not dying = 1-10^(-7)

P(30 die)=300000000C30 * (10^(-7))^30*(1-10^(-7))^299999970

P(30 die)=300000000C30 * 10^(-210)*(1-10^(-7))^299999970

The web2 calc won't do this calc for me but I saw a pop up of Alan's answer anyway :)

I think that it is the same as Alan's :/

Melody Aug 19, 2015