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# 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 di

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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

### Best Answer

#1
+26493
+10

Probability of an individual dying: p = 10-7

Probability of an individual surviving:  q = 1-p

Number of individuals: N = 3*108

Probability of exactly 30 individuals dying = (probability of 30 dying)*(probability of N-30 surviving)*(number of ways of choosing 30 out of N) = p30*qN-30*ncr(N,30) ≈ 0.073

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

.

Alan  Aug 19, 2015
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### 2+0 Answers

#1
+26493
+10
Best Answer

Probability of an individual dying: p = 10-7

Probability of an individual surviving:  q = 1-p

Number of individuals: N = 3*108

Probability of exactly 30 individuals dying = (probability of 30 dying)*(probability of N-30 surviving)*(number of ways of choosing 30 out of N) = p30*qN-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
+91793
+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

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