Creating a Probability Word Problem

[AmissArcher]

Hello, I'm trying to create a word problem, but it has been some time since I've worked on a problem like this and I am struggling to remember what formula to use. Specifically, I can't seem to figure out how I would find the probability of x amount of people not being a part of a group, given a big population.

(Problem)
In a city with a population of 18,900 people, there are 139 individuals that are over 90 years of age. If a group of 7 random individuals gathered together, what would be the odds that 1 or more of them were over 90 years of age?

(Solution)
Here is what I came up with, but not only does it feel inefficient, I can't shake the feeling that I did some "magic" math here.
139/18900 = 0.00735449735
P(x=>1)=P(x=1) + P(x=2) + ... + P(x=7)
P(x=1)= 0.00735449735
P(x=2)= 0.00735449735^2 = 0.00005408863
P(x=3)= 0.00735449735^3 = 0.00000039779
P(x=4)= 0.00735449735^4 = 0
P(x=5)= 0.00735449735^5 = 0
P(x=6)= 0.00735449735^6 = 0
P(x=7)= 0.00735449735^7 = 0 (these were all below e-9, so I rounded to zero)
The odds that 1 or more of the group mates were over 90 years old = 0.00740898377 = 0.7%

(Afterthought)
Since working through this some more, in attempts to make my mind easier to follow, would perhaps looking for P(x=>1) = 1 - P(x=0) be best? If so, how would I find P(x=0) ?

 

[Steven G]

Yes, it is best to use P(x=>1) = 1 - P(x=0).

How did you calculate the other probabilities (NOT odds!)? Can't you use the same method? Please post back showing how you did one of the probabilities you listed.

 

[HallsofIvy]

I am a bit disturbed that you immediately try to "remember what formula to use". Do you want to memorize formulas or learn mathematics?

You are told that "in a city with 18,900 people, there are 139 individuals that are over 90 years of age". You are asked to determine the probability that 1 or more out of 7 is over 90 years of age.
The probability "1 or more is 90 or over" is 1 minus the probability "all 7 are NOT over 90 years or age"
So we can just calculate the probability all 7 are not over 90 years of age then subtract that from 1.

There are two ways to do this. (That will give slightly different answers!)

The first method- and the really correct way to do it- is "choosing without replacement". Do you see that, since there are 139 people of 90 years of age or more, there are 18900- 139= 18761 who are NOT. So the probability the first person is not over 90 years of age is 18761/18900. Then there are 18899 people not yet chose, of whom 18760 are not over 90 years of age so the probability the next person is also not 90 years of age or more is 18760/18899. Continue to calculate that way for all 7 people and multiply the seven probabilities together, subtract from 1.

The second method- not really correct but much simpler and not far off- is "choosing with replacement". We imagine that we might well choose the same person again and count him or her as two of the seven! Since there are 18761 people who are not over 90 out of the 18899, the probability the first person chosen is, like before, 18760/18899= 0.9926. If we allow the possibility of choosing that same person again, the probability that all seven are not over 90 is \(\displaystyle (0.9926)^7\) and we can subtract that from 1.