Fast(er) way of computing the cumulative binomial probability? You flip a fair coin 100 times. What is the probability that you have less than 45 heads?

memLosycecyjz

memLosycecyjz

Answered question

2022-09-18

Fast(er) way of computing the cumulative binomial probability?
While revising for my probability test, I saw this question from one of the previous exams:
You flip a fair coin 100 times. What is the probability that you have less than 45 heads?
My question is a simple one. I know that to calculate the probability of a specific amount of n heads (or tails), we can use the binomial distribution with formula P ( X = k ) = ( n k ) p k ( 1 p ) n k , with n = 100 and p = 0.5 in this case. I also know that we can compute the chance of P ( X < k ) as either P ( X = 0 ) + P ( X = 1 ) + . . . + P ( X = k 10 ) or 1 ( P ( X = k ) + P ( X = k + 1 ) + . . . + P ( X = n ) )
However this is a simple question only worth two points out of over 60 total points. I cannot imagine that you need to compute P ( X = n ) 44 separate times and sum them together for such a small amount of points.
Is there any way to rewrite the formula or apply a different trick to drastically lower the amount of computations you need to do?

Answer & Explanation

Denier5h

Denier5h

Beginner2022-09-19Added 5 answers

Step 1
The trick is to indeed use the central limit theorem since we already have a large n.
Since our distribution is a binomial one, we have E [ X ] = μ = 0.5 100 = 50 and V a r ( X ) = σ 2 = 0.5 100 ( 1 0.5 ) = 25
Step 2
Then, we can compute 45 μ σ 2 = 45 50 25 = 1. Looking this up in the Z-table gives 0.1587. This is close to the exact answer of 0.14 (and indeed, both 0.14 and 0.16 are marked as correct in the answer sheet).

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