We are throwing darts on a disk-shaped board of radius 5. We assume that the proposition of the dart is a uniformly chosen point in the disk. The board has a disk-shaped bullseye with radius 1. Suppose that we throw a dart 2000 times at the board. Estimate the probability that we hit the bullseye at least 100 times.

the probability that we hit the bullseye at least 100 times is 0.0113

Step-by-step explanation:

Given the data in the question;

Binomial distribution

We find the probability of hitting the dart on the disk

⇒ Area of small disk / Area of bigger disk

⇒ πR₁² / πR₂²

given that; disk-shaped board of radius R² = 5, disk-shaped bullseye with radius R₁ = 1

so we substitute

⇒ π(1)² / π(5)² = π/π25 = 1/25 = 0.04

Since we have to hit the disk 2000 times, we represent the number of times the smaller disk ( BULLSEYE ) will be hit by X.

so

X ~ Bin( 2000, 0.04 )

n = 2000

p = 0.04

np = 2000 × 0.04 = 80

Using central limit theorem;

X ~ N( np, np( 1 - p ) )

we substitute

X ~ N( 80, 80( 1 - 0.04 ) )

X ~ N( 80, 80( 0.96 ) )

X ~ N( 80, 76.8 )

So, the probability that we hit the bullseye at least 100 times, P( X ≥ 100 ) will be;

we covert to standard normal variable

⇒ P( X ≥   )

⇒ P( X ≥ 2.28217 )

From standard normal distribution table

P( X ≥ 2.28217 ) = 0.0113

Therefore, the probability that we hit the bullseye at least 100 times is 0.0113

option: d is correct

step-by-step explanation: