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

Answer :

1234567822

Answer:

5% is the smallest probability

Step-by-step explanation:

5% = 0.05

2000 x 0.05 = 100 so the smallest chance is 5% but it could be more if you hit more bullseye

This is an example of a Binomial distribution, in which the chance of striking the dart here on the disc is:

[tex]=\frac{\text{area of the small disk} }{ \text{Area of the bigger disk}}[/tex]

[tex]= \frac{\pi\times R_1^2}{\pi \times R_2^2} \\\\ = \frac{1^2}{5^2} \\\\ = \frac{1}{25} \\\\ = 0.04[/tex]

Since this is a huge figure, one can use the central limit theory to approximate a supplied binomial distribution to a normally distributed one. As just a result, you have:

[tex]X \sim Bin(2000,0.04) n=2000 \ and\ p = 0.04[/tex]

Since this is  [tex]np = 2000 \times 0.04 = 80[/tex]  a huge figure, one can use the central limit theory to approximate a supplied binomial distribution to a normally distributed one. As just a result, you have:

[tex]X \sim N(np,np(1-p)) \\\\X \sim N(2000\times 0.04,2000 \times 0.04 \times (1-0.04)) \\\\X \sim N(80,76.8)[/tex]

We must now calculate the likelihood of hitting the bullseye at least 100 times:

[tex]P(X\geq 100)[/tex]

When we convert this to a conventional normal variable, we get:

[tex]P(Z \geq \frac{100-80}{\sqrt{76.8}})\\P(Z \geq 2.2822)[/tex]

When we use the conventional normal table to calculate this probability, we get:

[tex]P(Z \geq 2.2822) = 0.0112[/tex]

Therefore the required probability is "0.0112".

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