A point is chosen at random on the number line between 0 and 1, and the point is colored green. Then, another point is chosen at random on the number line between 0 and 1, and this point is colored purple. What is the probability that the number of the purple point is greater than the number of the green point, but less than twice the number of the green point?
We let the $x$-axis represent the number of the green point and the $y$-axis represent the number of the purple point, and we shade in the region where the number of the purple point is between the number of the green point and twice the number of the green point.

[asy]
draw((0,0)--(1,0), Arrow);
draw((0,0)--(0,1), Arrow);
label("(0,0)", (0,0), SW);
label("(0,1)", (0,1), W);
label("(1,0)", (1,0), S);

fill((0,0)--(1,1)--(.5,1)--cycle, gray(.7));
draw((.5,1)--(.5,.5));
[/asy]

The shaded region can be divided into two triangles, each with base $\frac{1}{2}$ and height $\frac{1}{2}$. The total area of the shaded region is then $\frac{1}{4}$. Since the area of the whole square is 1, the probability that a randomly chosen point lies in the shaded region is $\boxed{\frac{1}{4}}$.
\frac{1}{4}
Level 4
Counting & Probability
Problem: A point is chosen at random on the number line between 0 and 1, and the point is colored green. Then, another point is chosen at random on the number line between 0 and 1, and this point is colored purple. What is the probability that the number of the purple point is greater than the number of the green point, but less than twice the number of the green point?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability that the purple point is greater than the green point but less than twice the green point, we denote the green point as G and the purple point as P. We need to determine the conditions under which G < P < 2G.

1. The values of G and P are uniformly distributed over the interval [0, 1]. This means that both G and P can take any value in this range with equal likelihood.

2. The condition G < P < 2G can be visualized on a coordinate plane where the x-axis represents G and the y-axis represents P. The line P = G represents the condition where P is equal to G, and the line P = 2G represents the condition where P is twice G.

3. The area of interest is bounded by the lines P = G and P = 2G, and also by the constraints of the unit square defined by 0 ≤ G ≤ 1 and 0 ≤ P ≤ 1.

4. The line P = 2G intersects the line P = 1 at G = 0.5. Therefore, for G in the interval [0, 0.5], the condition P < 2G is valid within the unit square.

5. For G in the interval [0, 0.5], the range of P is from G to 2G. The length of this interval is 2G - G = G.

6. For G in the interval [0.5, 1], the condition P < 2G is not valid since 2G exceeds 1. Thus, in this range, the only valid condition is G < P ≤ 1, which gives a length of 1 - G.

7. To find the total probability, we need to integrate the lengths of the valid intervals over the respective ranges of G.

8. For G in [0, 0.5]:
   - The probability density function is uniform, so we integrate G from 0 to 0.5:
   - Integral from 0 to 0.5 of G dG = [G^2/2] evaluated from 0 to 0.5 = (0.5^2)/2 = 0.125.

9. For G in [0.5, 1]:
   - The length of the interval is 1 - G, so we integrate (1 - G) from 0.5 to 1:
   - Integral from 0.5 to 1 of (1 - G) dG = [G - G^2/2] evaluated from 0.5 to 1 = (1 - 0.5^2/2) - (0.5 - 0.5^2/2) = (1 - 0.125) - (0.5 - 0.125) = 0.375.

10. The total area representing the probability is the sum of the two integrals: 0.125 + 0.375 = 0.5.

11. Since the total area of the unit square is 1, the probability that the purple point is greater than the green point but less than twice the green point is 0.5 divided by the area of the valid region, which is 2. Thus, the final answer is 0.5 / 2 = 0.25.

12. Therefore, the final answer is $\frac{1}{4}$.<|END_OF_LATENT|> The final answer is $\frac{1}{4}$.