Pete's Probability Procedure 3

Let $x$ be the probability of getting a 2 or 5 with the dart, and let $y$ be the probability of getting a 1, 3, 4, or 6 with the dart. Then $2x + 4y = 1$ .

A player wins in the following cases: $\begin{array}{c|c|c} \text{dart} & \text{spinner} & \text{probability} \\ \hline 2 & 1 & \frac{x}{6} \\ 2 & 4 & \frac{x}{6} \\ 3 & 1 & \frac{y}{6} \\ 3 & 4 & \frac{y}{6} \\ 5 & 1 & \frac{x}{6} \\ 5 & 4 & \frac{x}{6} \\ 1 & 2 & \frac{y}{6} \\ 1 & 3 & \frac{y}{6} \\ 1 & 5 & \frac{y}{6} \\ 4 & 2 & \frac{y}{6} \\ 4 & 3 & \frac{y}{6} \\ 4 & 5 & \frac{y}{6} \end{array}$

Adding up the probabilities, we get that the overall probability of winning is $4 \cdot \frac{x}{6} + 8 \cdot \frac{y}{6} = \frac{4x + 8y}{6} = \frac{2(2x + 4y)}{6} = \frac{1}{3}.$

It is interesting to note that the probabilility distribution is different for the numbers in the case of spinner versus the case of the dart.

This can be understood in the following manner:

The spinner rotates about the point and is free to land in any direction, so it can be analysed in terms of the angle spanned by the shape at the vertex. Since all the angles are equal, this gives us: $P_s(1)=P_s(2)=P_s(3)=P_s(4)=P_s(5)=P_s(6)=\frac{1}{6}$ where $P_s(n)$ denotes the probability that the spinner will land on $n$ .

The dart, however has more freedom and this can intuitively understood by oberving that the region of $1, 3, 4, 6$ are bigger than the other ones. So, the probability here will be unequal and can be calculated by calculating the relative area of each region. If the side length of the square is $2$ , it can be seen that the area of the triangle containing $2$ is $\frac{1}{\sqrt{3}}$ , so the probability of the dart landing in region $2$ is $\frac{1}{4\sqrt{3}}$ . By symmetry and the fact that total probability is 1, it can be calculated that: $P_d(1)=P_d(3)=P_d(4)=P_d(6)=\frac{1}{4}-\frac{1}{8\sqrt{3}}, P_d(2)=P_d(5)=\frac{1}{4\sqrt{3}}$ where $P_d(n)$ denotes the probability that the spinner will land on $n$ .

Since the questions states that it doesn't matter which out of the spinner or dart got the prime or square, just one of each is needed, we have two cases:

1. Spinner lands on prime, dart lands on square.
2. Spinner lands on square, dart lands on prime.

(Since no number is both a square and a prime, we need not worry about any other case.) Primes: 2, 3, 5 Squares: 1, 4

$\therefore \text{Required probability=} (P_s(2)+P_s(3)+P_s(5))(P_d(1)+P_d(4)) + (P_d(2)+P_d(3)+P_d(5))(P_s(1)+P_s(4))$

$=(\frac{1}{2})(\frac{1}{2}-\frac{1}{4\sqrt{3}}) + (\frac{1}{3})(\frac{1}{4}+\frac{3}{8\sqrt{3}})$

$=\frac{1}{4}+\frac{1}{12}$

$=\boxed{\frac{1}{3}}$