Suppose a circular dart board is $18$ inches in diameter, and the
bull’s eye in the center is $2$ inches in diameter. What is the probability that out of
four darts thrown at the dart board, exactly two hit the bull’s eye? Assume that
each dart lands at a uniformly random point on the board, independently of the
other darts.
I am just wondering if my method was correct, I found that the total area was $81\pi$ and the area of the center was $4\pi$, then I divide $\frac{4\pi}{81\pi}$ and then squared it to get $.0024$. Is that correct? and if it is not, Can you show where I went wrong.
Thanks.
Best Answer
Your process started out correct: The area of the board is
$$A_b=\pi r^2=81\pi$$
but the area of the bulls-eye is
$$A_e=\pi r^2=\pi$$
So the probability that any dart hits the bulls-eye is
$$P=\frac{A_e}{A_b}=\frac{\pi}{81\pi}=\frac{1}{81}$$
Now, to find the probability that exactly two darts hits the bulls-eye, first note there are exactly $\binom{4}{2}$ (read $4$ choose $2$) ways to choose exactly two darts from the four. This is defined as
$$\binom{4}{2}=\frac{4!}{2!(4-2)!}=\frac{24}{2\cdot 2}=6$$
Second, the probability that exactly two darts (in any order) hit the center is given by
$$P\cdot P\cdot (1-P)\cdot (1-P)=\frac{6400}{43046721}$$
Since there six ways to find arrange the darts and have this happen, the final probability of exactly two darts hitting the bulls-eye is given by
$$6\frac{6400}{43046721}=\frac{12800}{14348907}=0.0892054\%$$