find the sum of P(8), P(9), P(10)

Question

to find the probability that the number who consider themselves baseball fans is at least eight, find the sum of P(8), P(9),P(10)

Answer 1

I need more info, so I'm assuming there are 10 people here. Each considers himself/herself a baseball fan or not a baseball fan. Let's say there's an equal probability of someone being a baseball fan or of not being a baseball fan, so p=½. Consider the equation (½+½)ⁿ=1. No matter what the value of n is this equation will always be true because 1ⁿ=1. So let n=10: (½+½)¹⁰=1. This is a binomial probability distribution and we can expand this (without evaluating it):

½¹⁰+10(½)⁹(½)+45(½)⁸(½)²+120(½)⁷(½)³+210(½)⁶(½)⁴+...

The coefficients can be calculated thus:

10, (10×9)/(1×2)=45, (10×9×8)/(1×2×3)=120, (10×9×8×7)/(1×2×3×4)=210, and so on.

The list of coefficients in order is:

1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1 (11th row of Pascal's Triangle).

Note that pⁿp^10-n=p¹⁰ for every term.

The sum of the series is 1 because (½+½)¹⁰=1.  So ½¹⁰(1+10+45+120+210+...)=1.

When we add the coefficients together we get 1024=2¹⁰.

We are interested in the sum of the last three or the first three terms (because of the symmetry):

(1+10+45)(½¹⁰)=56/1024=7/128=0.0547 approx or 5.47%. This is P(8)+P(9)+P(10)=P(1)+P(2)+P(3) (in this case of equal probability, p=½). 

But, suppose that, on average, in the population as a whole, 4 out of 5 people are baseball fans then 1 out of 5 people is not a baseball fan. This alters the balance and we have to consider something different, but similar, in most ways to what we just calculated. In this case though, p=⅘ and 1-p=⅕. Let q=1-p=⅕.

Now we have:

p¹⁰+10p⁹q+45p⁸q²+120p⁷q³+... 

The first 3 terms are P(10), P(9), P(8), which is the probability that at least 8 people out of 10 randomly chosen are baseball fans.

We would have:

P(10)+P(9)+P(8)=p¹⁰+10p⁹q+45p⁸q²=⅘¹⁰+10(⅘)⁹(⅕)+45(⅘)⁸(⅕)²=0.6778=67.78%.