# Probability of throwing darts when likelyhood of hitting one side is more than the other

Currently learning probability from MIT OCW 6.041. I want to understand the following regarding Problem #5 from the PSET1(read the question here at MIT OCW). I am unable to understand part c of the given problem.

We need to calculate the probability of John hitting 50 point mark on the dartboard. I understand that we are looking for values of $$\alpha$$ and $$\beta$$ which are the proportionality constants, I want to understand why the probability of throwing the dart in the right half of the board is $$\frac{2}{3}$$

The question being asked above(in bold) is being referenced from the answer given for the Problem #5 referenced from the solution of the PSET1-Solution from the MIT OCW site.

The first term in the sum is proportional to the area of the right half of the inner disk and is equal to απR2/2 = απ/2, where α is a constant to be determined. Similarly, the probability of him throwing in the left half of the board and scoring 50 points is equal to βπ/2, where β is a constant (not necessarily equal to α). In order to determine α and β, let us compute the probability of throwing the dart in the right half of the board. This probability is equal to απR2/2 = απ102/2 = α50π. Since that probability is equal to 2/3, α = 1/(75π). In a similar fashion, β can be determined to be 1/(150π). Consequently, the total probability is equal to 1/150 + 1/300 = 0.01

I want to understand why the probability of throwing the dart in the right half of the board is $$\frac{2}{3}$$

We are told that "John is right handed and is twice more likely to throw in the right half of the board than in the left half."

So, $$P(\text{Right half of the board}) = 2P(\text{Left half of the board})$$.

Since there are only two sides of the board, $$P(\text{Right half of the board}) + P(\text{Left half of the board}) = 1 = P(\text{hit on the board})$$ $$\text{(we're gonna assume missing the board is impossible)}$$

Substituting in our first equation,

$$2P(\text{Left half of the board}) + P(\text{Left half of the board}) = 1$$ $$3P(\text{Left half of the board}) = 1$$ $$P(\text{Left half of the board}) = \frac{1}{3}$$ $$P(\text{Right half of the board}) = 2P(\text{Left half of the board}) = \frac{2}{3}$$

Since John is twice as likely to hit the right part of the board, for every one dart that he lands on the left half, he would have landed two darts on the right half.

Thus out of every 3 darts thrown, two land on the right half and one lands on the left half. Thus $$\alpha = \frac{2}{3}$$ and $$\beta = \frac{1}{3}$$.