The mosquito is fixed to a straight line. Each day, it can choose to whether to fly a kilometer to the northeast or a kilometer to the southwest. It is unbiased, so each path is equally likely. The claim is that. The chance that a mosquito on its two-hundredth day of life is at least 40km from home is just under 3 in 1000.

sbrigynt7b

sbrigynt7b

Answered question

2022-11-19

I'm reading through Shape by Jordan Ellenberg and came across this claim, modeling the movement of a mosquito as a binomial distribution.
The mosquito is fixed to a straight line. Each day, it can choose to whether to fly a kilometer to the northeast or a kilometer to the southwest. It is unbiased, so each path is equally likely.
The claim is that
The chance that a mosquito on its two-hundredth day of life is at least 40km from home is just under 3 in 1000.
A footnote adds that the exact computation is
"What is the probability that a binomial random variable with p = 0.5 and n = 200 takes value at least 120?"
To cover a distance of, say 40 km northeast, the mosquito would need to have moved 120 km northeast and 80 km southwest. Representing the movement northeast as a "success", this could be the binomial distribution described above.
X B i n ( 0.5 , 200 )
And that we are looking to calculate P ( X 120 )
However, it strikes me that the chance that the mosquito is at least 40km from home would be twice of this computed value, since it could be a net 40km northeast or a net 40km southwest. Am I misunderstanding the claim above or modeling the probability incorrectly?

Answer & Explanation

Marshall Flowers

Marshall Flowers

Beginner2022-11-20Added 20 answers

Step 1
An unambiguous way to model the problem would be to consider the IID random variables
X 1 , X 2 , , X 200 { 1 , 1 }
where
Pr [ X i = 1 ] = Pr [ X i = 1 ] = 1 2 .
Then the distance of the mosquito from the origin after 200 days will simply be
D = | i = 1 200 X i | .
To transform this into a binomial model, we note that
X i + 1 2 Bernoulli ( p = 1 / 2 ) .
Therefore
D = | i = 1 200 ( 2 X i + 1 2 1 ) | = 2 | i = 1 200 X i + 1 2 100 | = 2 | Y 100 | ,
where
Y Binomial ( n = 200 , p = 1 / 2 ) .
It follows that
Pr [ D 40 ] = Pr [ 2 | Y 100 | 40 ] = Pr [ | Y 100 | 20 ] = 1 Pr [ | Y 100 | < 20 ] = 1 Pr [ 20 < Y 100 < 20 ] = 1 Pr [ 80 < Y < 120 ] = 1 y = 81 119 ( 200 y ) 2 200 = 570980841170325361571226368804554522130683181662812200441 100433627766186892221372 630771322662657637687111424552206336 0.00568516.
This is twice the claimed probability.

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