statistics - Calculate probability of an event not by exclusion

Question

I have some doubt with these kind of problems, example:

"If we asked 20,000 in a stadium to toss a coin 10 times, what it's the probability of at least one person getting 10 heads?"

I took this example from Practical Statistics for Data Scientist.

So, the probability of at least one person getting 10 heads it's calculated using: 1 - P(of nobody in the stadium getting 10 heads).

So we kind of doing an exclude procedure here, first I get the probability of the contrary event I am trying to measure, not the ACTUAL experiment I want to measure: at least one people getting 10 heads.

Why do we do it this way?

How can I calculate the probability of at least someone getting 10 heads but without passing through the probability of no one getting 10 heads?

question from:https://stackoverflow.com/questions/65894777/calculate-probability-of-an-event-not-by-exclusion

Answer 1

As @Robert Dodier mentioned in the comments, the reason is that the calculations are simpler. I will use a stadium of 20 people instead of 20000 as an example:

Method 1:

Probability of not getting 10 heads for one individual

= 1 - probability of getting 10 heads
= 1 - 10!/(10!0!)*0.5^10*(1-0.5)^0
= 0.9990234375

Probability of at least one person in the stadium getting 10 heads

= 1 - P(of nobody in the stadium getting 10 heads)
= 1 - 0.9990234375**20 (because all coin tosses are independent)
= 0.019351109194852834

Method 2:

Probability of getting 10 heads for one individual

= 10!/(10!0!)*0.5^10*(1-0.5)^0
= 0.0009765625

Probability of exactly 1, 2, 3, etc. persons in the stadium getting 10 heads:

p1 = 20!/(1!19!)*0.0009765625^1*(1-0.0009765625)^(20-1) = 0.019172021325613825
p2 = 20!/(2!18!)*0.0009765625^2*(1-0.0009765625)^(20-2) = 0.00017803929872270904
p3 = 20!/(3!17!)*0.0009765625^3*(1-0.0009765625)^(20-3) = 1.0442187608370032e-06
p4 = 20!/(4!16!)*0.0009765625^4*(1-0.0009765625)^(20-4) = 4.338152232216289e-09
p5 = 20!/(5!15!)*0.0009765625^5*(1-0.0009765625)^(20-5) = 1.3569977656981548e-11
p6 = 20!/(6!14!)*0.0009765625^6*(1-0.0009765625)^(20-6) = 3.316221323798032e-14
p7 = 20!/(7!13!)*0.0009765625^7*(1-0.0009765625)^(20-7) = 6.483326146232712e-17
p8 = 20!/(8!12!)*0.0009765625^8*(1-0.0009765625)^(20-8) = 1.029853859983202e-19
p9 = 20!/(9!11!)*0.0009765625^9*(1-0.0009765625)^(20-9) = 1.342266353839299e-22 
p10 = 20!/(10!10!)*0.0009765625^10*(1-0.0009765625)^(20-10) = 1.443297154665913e-25
p11 = 20!/(11!9!)*0.0009765625^11*(1-0.0009765625)^(20-11) = 1.2825887804726853e-28
p12 = 20!/(12!8!)*0.0009765625^12*(1-0.0009765625)^(20-12) = 9.403143551852531e-32
p13 = 20!/(13!7!)*0.0009765625^13*(1-0.0009765625)^(20-13) = 5.656451493707817e-35
p14 = 20!/(14!6!)*0.0009765625^14*(1-0.0009765625)^(20-14) = 2.7646390487330485e-38
p15 = 20!/(15!5!)*0.0009765625^15*(1-0.0009765625)^(20-15) = 1.0809927854283668e-41
p16 = 20!/(16!4!)*0.0009765625^16*(1-0.0009765625)^(20-16) = 3.3021529369146104e-45
p17 = 20!/(17!3!)*0.0009765625^17*(1-0.0009765625)^(20-17) = 7.59508466888531e-49
p18 = 20!/(18!2!)*0.0009765625^18*(1-0.0009765625)^(20-18) = 1.2373875315877011e-52
p19 = 20!/(19!1!)*0.0009765625^19*(1-0.0009765625)^(20-19) = 1.2732289258503896e-56
p20 = 20!/(20!0!)*0.0009765625^20*(1-0.0009765625)^(20-20) = 6.223015277861142e-61
 

Probability of at least one person in the stadium getting 10 heads

= p1 + p2 + p3 + p4 + p5 + p6 + p7 + p8 + p9 + p10 + 
p11 + p12 + p13 + p14 + p15 + p16 + p17 + p18 + p19 + p20
= 0.01935110919485281
 

So the result is the same (the tiny difference is due to floating point precision), but as you can see the first calculation is slightly simpler for 20 people, never mind for 20000 ;)